Suppose a vending machine's receipt says the price was 3 dollars, and you want to know which buttons could have produced that price. You are searching backward from an output to possible inputs. To solve a function equation means find every allowed input that gives the stated output. This differs from evaluation, where the input is already known. A function promises one output for each input, but several inputs may share that output. Your search might find two inputs, one input, or none. Substituting each answer back into the original recipe checks whether it really produces the requested result.
Solving for output 3 finds both inputs pointing to it.
Reminder
Equal operations on both sides. p + 3 = 0 becomes p = −3 by subtracting 3 from both sides.
Why it works. The equation h(p) = 3 means that the output h(p) equals 3, so replace h(p) with its formula and find p. Equal changes to both sides preserve equality, like removing equal weights from a balance. For a quadratic, rewriting as a product equal to zero makes the search manageable. A product can be zero only if at least one factor is zero: two nonzero factors cannot multiply to zero. Each factor then gives a smaller equation. Check every resulting input in the original formula.
RuleTo solve f(x) = k, set the function's formula equal to k and find every allowed input x. Zero product rule: AB = 0 means A = 0 or B = 0, or both.
The same idea, five ways
Say it
Solve h of p equals three
Write it
Find every input p whose output is 3.
In math
h(p) = 3
p2 + 2p = 3
p = −3 or p = 1
Like
A receipt gives the price; find every matching button.
See itThe requested output 3 occurs at two different input positions.
The same idea, other ways
Search backward through a machine
Evaluation gives the button and asks for the receipt. Solving gives the receipt and asks for every matching button. For h(p) = p2 + 2p, input 1 and input −3 both produce 3, so both belong in the answer.
Two different inputs can share the requested output without breaking the function.
Keep a scale balanced
If 3x − 7 = 8, adding 7 to both sides gives 3x = 15, and dividing both sides by 3 gives x = 5. You preserve the balance while uncovering the unknown input.
Apply the same operation to both sides while finding the input.
Split a zero product into two doors
If (p + 3)(p − 1) = 0, the first factor can be zero or the second can be zero. These possibilities give p = −3 and p = 1. Keep both doors open until you check each candidate in the original function.
(p + 3)(p − 1) = 0
p + 3 = 0 or p − 1 = 0
p = −3 or p = 1
One product equation becomes two possible input equations.
.1One rule, two directions
Use the same recipe to move forward from an input or backward from an output. The name h labels the rule h(p) = p2 + 2p. Here h is the function name and p is its input. In the earlier difference quotient, h instead named a numerical step. A letter gets its meaning from its definition. First evaluate h(4), then solve the output question h(p) = 3 in the main example.
h(4) gives input 4 and asks for output.
h(p) = 3 gives output 3 and asks for every matching input.
The function is named h here; its input variable is p.
The same idea, five ways
Say it
h of four asks for an output; h of p equals three asks for inputs.
Write it
Evaluation starts with input 4, while solving starts with the specified output 3.
In math
h(p) = p2 + 2p
h(4) = 24
h(p) = 3
p = −3 or p = 1
Like
A button finds its receipt; a receipt may identify several matching buttons.
See itThe function is named h here; its input variable is p.
Worked exampleEvaluate a rule whose input letter is p
For h(p) = p2 + 2p, evaluate h(4). The input is 4; find the output.
The function is named h here; its input variable is p.
What it asks. Use each stated input to find its output from the named formula.
Plan. Replace every input-variable slot with the whole input in parentheses. Work powers before multiplication, and multiplication before adding or subtracting. For a square root, compute its inside first and choose the nonnegative root. Check by following the original recipe again.
Every input slot receives the complete input, including its sign.
Square a negative input in parentheses: (−2)2 = 4. A constant recipe gives the same output for every input.
Keep separate requested outputs on separate answer lines.
h(4) = (4)2 + 2(4) = 16 + 8 = 24.The input letter is p in this new formula. Replace every p with 4, then square and multiply before adding.
Answer
h(4) = 24.
Check Compute the two parts separately: 4 × 4 = 16 and 2 × 4 = 8. Their sum is 24.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: h(p) = 3 means the input p equals 3.
The equals sign sets the function output to 3.
✓ Instead: Set p2 + 2p = 3 and find the matching p values.
Tips and tricks
Read the entire question before deciding its direction.
As an everyday comparison: Use the same recipe to move forward from an input or backward from an output. The name h labels the rule h(p) = p2 + 2p. Here h is the function name and p is its input. In the earlier difference quotient, h instead named a numerical step. A letter gets its meaning from its definition. First evaluate h(4), then solve the output question h(p) = 3 in the main example.
With the worked values: Compute the two parts separately: 4 × 4 = 16 and 2 × 4 = 8. Their sum is 24.
.2Balance a linear equation
A linear recipe multiplies the input by a fixed number, then adds or subtracts a fixed number. Think of a price made from a fixed fee and a charge per item. To find the number of items from the total, undo the fixed fee first, then undo the per-item charge. An equation acts like a balance scale. Whatever you add, subtract, multiply or divide on one side must also happen on the other, so the two amounts stay equal.
An equation states that its two sides have the same value.
Adding or subtracting the same amount on both sides preserves equality.
Dividing both sides by the same nonzero number preserves equality.
Adding 7 to each side exposes the tripled input.
Reminder
Canceling a nonzero factor.3x3 = x because 3 multiplies all of the numerator and 3 ≠ 0.
The same idea, five ways
Say it
Three x minus seven equals eight, so x equals five.
Write it
Undo the last operation first and do the same operation on both sides of the equation.
In math
3x − 7 = 8
3x = 15
x = 5
3(5) − 7 = 8
Like
Keep both pans of a scale balanced while removing equal amounts.
See itAdding 7 to each side exposes the tripled input.
Worked exampleFind the input for output 8
For r(x) = 3x − 7, solve r(x) = 8. The output is 8; find the input.
3x − 7 = 8.
3x − 7 + 7 = 8 + 7, so 3x = 15.
3x3 = 153, so x = 5.
Read these successive steps along with their reasons.
What it asks. The output is given. Find every allowed input that produces it.
Plan. Replace the function call by its formula. Set that formula equal to the requested output to turn the backward question into an equation. Undo the added or subtracted constant on both sides, then divide both sides by the nonzero input coefficient. Check the recovered input in the original formula.
The value after the equals sign is the requested output.
Use the same operation on both sides to preserve equality.
Substitution in the starting formula must return the target output.
3x − 7 = 8.Replace r(x) with the function's formula.
3x − 7 + 7 = 8 + 7, so 3x = 15.Add 7 to both sides to undo the subtraction of 7.
3x3 = 153, so x = 5.Divide both sides by the nonzero factor 3.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: From 3x − 7 = 8, write 3x = 8 − 7 = 1.
Subtracting 7 does not undo the existing subtraction of 7. Adding 7 to both sides does.
✓ Instead: 3x = 8 + 7 = 15, so x = 5.
Tips and tricks
Write the operation on both sides once, so a sign mistake is visible.
Undo a recipe in reverse: The recipe 3x − 7 triples first and subtracts 7 second. Undo it by adding 7 first and dividing by 3 second.
Equal changes preserve equal totals: Two equal piles stay equal when each receives seven more pieces. Two equal totals also stay equal when each is divided into three equal groups.
.3Factor and use the zero product rule
Factoring repacks a sum as a multiplication, like putting loose items back into equal bags. A factor is a quantity being multiplied. For a quadratic beginning with x2, look for two numbers whose product gives the constant term and whose sum gives the coefficient of x. Their two parentheses multiply back to the original expression. This becomes useful for solving after the equation equals zero, because a zero product tells you that at least one whole factor must be zero.
The constant term is the number without the variable. In x2 + 3x − 10, that term is −10.
For x2 + bx + c, if r + s = b and rs = c, then x2 + bx + c = (x + r)(x + s).
The zero product rule applies to a product equal to zero. First move the requested output to the other side.
The four products of (x + 5)(x − 2) combine to x2 + 3x − 10; negative labels represent algebraic products.
Reminder
Distributive property. (x + 3)(x − 1) gives four products: x2 − x + 3x − 3.
Solving a small equation. p + 3 = 0 gives p = −3 after subtracting 3 from both sides.
The same idea, five ways
Say it
x squared plus three x minus ten equals the quantity x plus five times the quantity x minus two.
Write it
Factoring repacks the quadratic sum as a product; a product equal to zero has at least one zero factor.
In math
5 + (−2) = 3
5 × (−2) = −10
x2 + 3x − 10 = (x + 5)(x − 2)
(x + 5)(x − 2) = 0
x = −5 or x = 2
Like
Repack loose items into multiplied groups, then find which group count can be zero.
See itThe four products of (x + 5)(x − 2) combine to x2 + 3x − 10; negative labels represent algebraic products.
Worked exampleSolve f(x) = 6 by factoring
For f(x) = x2 + 3x − 4, solve f(x) = 6. The output is 6; find every input giving that output.
The level line at output 6 meets the curve at inputs −5 and 2.
What it asks. The output is given. Find every allowed input that produces it.
Plan. Set the formula equal to the requested output. If there is a middle term, subtract the output to make one side zero, factor, and set each factor to zero. If only the square remains, keep both root signs when the right side is positive. Check every candidate in the starting formula.
An output after the equals sign is the target, not an input to substitute.
The zero product rule requires the entire product to equal zero.
Keep every candidate and check each one in the original formula.
x2 + 3x − 4 = 6.The formula must equal the requested output.
x2 + 3x − 4 − 6 = 0, so x2 + 3x − 10 = 0.Subtract 6 from both sides to make one side zero. Zero makes the zero product rule applicable after factoring, so this step prepares a way to find every matching input.
List pairs with product −10 and their sums: 1 and −10 give −9; −1 and 10 give 9; 2 and −5 give −3; −2 and 5 give 3. Choose −2 and 5.The two factor numbers must reproduce the constant −10 and the middle coefficient 3.
x + 5 = 0 or x − 2 = 0.At least one factor must be zero for their product to be zero.
x = −5 or x = 2.Subtract 5 in the first equation and add 2 in the second.
Answer
x = −5
x = 2
Check f(−5) = (−5)2 + 3(−5) − 4 = 25 − 15 − 4 = 6. Also f(2) = 4 + 6 − 4 = 6. Both inputs give the requested output.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Choose numbers that multiply to the constant without checking their sum.
For x2 + 2x − 3, choosing −3 and 1 gives the right product −3 but the wrong sum −2.
✓ Instead: Choose 3 and −1, giving both product −3 and sum 2.
✗ Not this: (x − 1)(x + 2) = 4 gives x = 1 or x = −2.
Those values make the product 0, but the required product is 4. The zero product rule requires zero on the other side.
✓ Instead: Expand to x2 + x − 2 = 4. Subtract 4 to get x2 + x − 6 = 0, which makes the zero product rule usable. Factor as (x + 3)(x − 2) = 0, giving x = −3 or x = 2. Check: (−4)(−1) = 4 and 1 × 4 = 4.
Tips and tricks
Write the required sum and product beside the quadratic. Expand your factors once to check both.
Rebuild the middle and last terms: In (x + r)(x + s), the two middle products combine to (r + s)x, while the last product is rs. Matching the sum and product makes the multiplication reproduce your quadratic.
One zero factor stops the product: If either bag count in a multiplication is zero, the total is zero. If both factors are nonzero, their product stays nonzero. That is why checking each factor finds every way a factored quadratic can equal zero.
.4Practice the factoring search before solving
The sum-and-product rule tells you what to look for. To find the pair, list whole-number pairs of the constant and test their sums. Matching signs give a positive product. Opposite signs give a negative product. If both numbers must be negative, start with positive factor pairs and reverse both signs. Expand the result once to check it. The five factoring rungs below show this search before the solving rungs begin.
If the constant is positive, the two numbers have the same sign, chosen to match the middle sum.
If the constant is negative, the signs differ; the number with larger absolute value has the sign of the middle sum.
A zero constant gives an immediate common factor x, as in x2 + 4x = x(x + 4).
If the middle coefficient is zero and the constant is negative, the matching numbers have equal size and opposite signs. For example, x2 − 16 = (x − 4)(x + 4), because −4 + 4 = 0 and (−4) × 4 = −16.
Both negative constants give a positive last product and a negative middle sum.
The same idea, five ways
Say it
Negative four plus negative five equals negative nine, and their product is twenty.
Write it
Choose factor numbers whose sum matches the middle coefficient and whose product matches the constant.
In math
−4 + (−5) = −9
(−4)(−5) = 20
x2 − 9x + 20 = (x − 4)(x − 5)
Like
A pair must fit both locks: the required sum and the required product.
See itBoth negative constants give a positive last product and a negative middle sum.
Worked exampleFactoring rung 5: two negative numbers
Factor x2 − 9x + 20. Find two parentheses that multiply back to this sum.
Both negative constants give a positive last product and a negative middle sum.
What it asks. Rewrite the same sum as multiplied factors. Do not change its value.
Plan. Identify the constant and middle coefficient. List integer factor pairs of the constant, check their sums and signs, and write the pair in parentheses. Multiply the factors back to check.
The two constant numbers must multiply to the last term.
Their sum must equal the coefficient of the middle term.
Check the signs, then multiply all four term pairs back.
The required sum is −9 and product is 20.The middle coefficient and constant determine both targets.
The positive pairs of 20 are 1 and 20, 2 and 10, 4 and 5. Their sums are 21, 12, and 9.Listing pairs prevents guessing a pair that matches only the product.
Use −4 and −5: their sum is −9 and their product is 20.A positive product and negative sum require both numbers to be negative.
x2 − 9x + 20 = (x − 4)(x − 5).The constants are the pair matching both requirements.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this:x2 − 9x + 20 = (x + 4)(x + 5).
The product 20 matches but the sum is 9 rather than −9.
✓ Instead: Use (x − 4)(x − 5).
Tips and tricks
Write both targets first: sum and product.
As an everyday comparison: The sum-and-product rule tells you what to look for. To find the pair, list whole-number pairs of the constant and test their sums. Matching signs give a positive product. Opposite signs give a negative product. If both numbers must be negative, start with positive factor pairs and reverse both signs. Expand the result once to check it. The five factoring rungs below show this search before the solving rungs begin.
With the worked values: Expansion gives x2 − 5x − 4x + 20 = x2 − 9x + 20.
.5Factoring the opposite-sign quadratic
For x2 + 2x − 3, a negative product requires one positive number and one negative number. The pair 3 and −1 gives the required positive sum 2. This is the same factor pattern used in the p equation. The variable’s name changes the label, not the multiplication.
Check both the sum and product before writing the parentheses.
The two middle products −x and 3x combine to 2x.
The same idea, five ways
Say it
x squared plus two x minus three equals the quantity x plus three times the quantity x minus one.
Write it
Opposite signs give the negative constant, while the larger positive number gives the positive middle sum.
In math
3 + (−1) = 2
3 × (−1) = −3
x2 + 2x − 3 = (x + 3)(x − 1)
p2 + 2p − 3 = (p + 3)(p − 1)
Like
Two number tags must rebuild both the middle amount and the final amount.
See itThe two middle products −x and 3x combine to 2x.
Worked exampleFactoring rung 4: the quadratic
Factor x2 + 2x − 3. This is the same pattern used when solving h(p) = 3.
The two middle products −x and 3x combine to 2x.
What it asks. Rewrite the same sum as multiplied factors. Do not change its value.
Plan. Identify the constant and middle coefficient. List integer factor pairs of the constant, check their sums and signs, and write the pair in parentheses. Multiply the factors back to check.
The two constant numbers must multiply to the last term.
Their sum must equal the coefficient of the middle term.
Check the signs, then multiply all four term pairs back.
The desired sum is 2 and the desired product is −3.The coefficient of x is 2 and the constant is −3.
Choose 3 and −1, since 3 + (−1) = 2 and 3 × (−1) = −3.A correct pair must satisfy both conditions.
x2 + 2x − 3 = (x + 3)(x − 1).Place the two numbers into the two factors.
Answer
x2 + 2x − 3 = (x + 3)(x − 1)
Check Expand: (x + 3)(x − 1) = x2 − x + 3x − 3 = x2 + 2x − 3. Replacing x by p gives p2 + 2p − 3 = (p + 3)(p − 1).
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Use −3 and 1 because their product is −3.
Their sum is −2, so the middle term is wrong.
✓ Instead: Use 3 and −1, whose sum is 2.
Tips and tricks
Changing x to p changes every occurrence together.
As an everyday comparison: For x2 + 2x − 3, a negative product requires one positive number and one negative number. The pair 3 and −1 gives the required positive sum 2. This is the same factor pattern used in the p equation. The variable’s name changes the label, not the multiplication.
With the worked values: Expand: (x + 3)(x − 1) = x2 − x + 3x − 3 = x2 + 2x − 3. Replacing x by p gives p2 + 2p − 3 = (p + 3)(p − 1).
.6Solving an equation with a square root
A square root is like asking for the side length of a square of a known area. It returns one nonnegative length. When its output is given, square that output to recover the inside, then solve the remaining equation. Check the requested sign first. A negative requested output cannot come from this square-root symbol. After squaring, always put your input back into the starting root formula to check it.
t − 6 requires t − 6 ≥ 0, so t ≥ 6. This finds the inputs with a nonnegative inside; at t = 6 the root is 0 = 0.
For a requested nonnegative output, squaring both sides removes the square-root symbol.
Checking in the starting equation rejects candidates created by squaring.
The curve gives output 2 at input 10 and output 7 at input 55.
Reminder
Principal square root.4 = 2, while solving z2 = 4 gives z = 2 or z = −2.
The same idea, five ways
Say it
The square root of the whole difference t minus six equals seven, so t equals fifty-five.
Write it
A nonnegative root output can be squared to recover the inside; check the recovered input in the starting root formula.
In math
k(t) = t − 6
t − 6 ≥ 0
t ≥ 6
t − 6 = 7
t − 6 = 49
t = 55
55 − 6 = 7
Like
A square’s side length recovers its area when you square that length.
See itThe curve gives output 2 at input 10 and output 7 at input 55.
Worked exampleEvaluate and solve a shifted square root
Let k(t) = t − 6. Find k(10), then solve k(t) = 7. The first gives an input and wants an output. The second gives an output and wants every input.
Subtract 6 before taking the nonnegative root.
What it asks. For the stated input, find its output first. Then find every allowed input giving the requested output.
Plan. For any requested evaluation, compute the inside before taking its nonnegative root. For solving, check that the requested root output is nonnegative, square both sides, isolate the input, and check it in the starting root equation.
The principal square root returns a nonnegative value.
Squaring both sides recovers the expression inside the root.
A candidate must pass the original root equation, including its sign.
k(10) = 10 − 6 = 4 = 2.Replace t by 10 and take the nonnegative square root.
To solve, write t − 6 = 7.Set the whole output formula equal to the requested nonnegative output.
Square both sides: t − 6 = 49.For a nonnegative square root, squaring removes the root; 72 = 49.
Add 6 on both sides to get t = 55.This isolates the input whose inside quantity must be 49.
Check k(55) = 55 − 6 = 49 = 7.Squaring can hide an original sign requirement, so check in the original root formula.
Answer
k(10) = 2.
t = 55.
Check The output 7 squares to the inside value 49. At input 55, the inside is 49, returning the required nonnegative 7.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Solving k(t) = −2 by squaring gives t = 10, so 10 is a solution.
The principal root cannot be negative. The candidate returns k(10) = 2, which is not −2.
✓ Instead: There is no solution to k(t) = −2.
Tips and tricks
Check the requested output sign before squaring, then substitute back afterward.
As an everyday comparison: A square root is like asking for the side length of a square of a known area. It returns one nonnegative length. When its output is given, square that output to recover the inside, then solve the remaining equation. Check the requested sign first. A negative requested output cannot come from this square-root symbol. After squaring, always put your input back into the starting root formula to check it.
With the worked values: The output 7 squares to the inside value 49. At input 55, the inside is 49, returning the required nonnegative 7.
Strategy: step by step
Say the question in words. Solve h(p) = 3 means the output is 3; find every input p that produces it.
Replace the function notation with its formula, keeping the requested output on the other side.
For a linear equation, undo addition or subtraction and then undo multiplication or division, doing each operation to both sides.
For a quadratic that factors, subtract the requested output from both sides so one side is zero.
Rewrite the quadratic as a product of factors. Set each factor equal to zero using the zero product rule.
Solve each smaller equation and keep every permitted input.
Substitute each answer into the original function. Each checked output must equal the requested output.
Strategy
Choose the solving method for this formula
1
Is the formula linear?
YesUndo addition or subtraction, then divide by the nonzero multiplier.
NoIdentify the power, root, or other operation.
↓
2
Is the equation a square equal to a number?
YesA positive number gives both square-root signs, zero gives one solution, and a negative number gives no real solution.
NoIf it is a factorable quadratic with a middle term, make one side zero, then factor.
↓
3
Is the equation a square root alone equal to a number, inside = k?
YesA negative k has no solution. For a nonnegative k, square both sides, solve, and check. An outside minus or added term must be handled first.
NoUse the operation actually shown and check every resulting input.
Say the question in words. Solve h(p) = 3 means the output is 3; find every input p that produces it.
Replace the function notation with its formula, keeping the requested output on the other side.
For a linear equation, undo addition or subtraction and then undo multiplication or division, doing each operation to both sides.
For a quadratic that factors, subtract the requested output from both sides so one side is zero.
Rewrite the quadratic as a product of factors. Set each factor equal to zero using the zero product rule.
Solve each smaller equation and keep every permitted input.
Substitute each answer into the original function. Each checked output must equal the requested output.
Worked exampleTwo inputs for output 3
Let h(p) = p2 + 2p. Solve h(p) = 3. The output is 3; find every input p that gives it. Here h names the function, unlike the change variable h in the previous lesson.
The requested output 3 occurs at two different input positions.
What it asks. The output is given. Find every allowed input that produces it.
Plan. Set the formula equal to the requested output. If there is a middle term, subtract the output to make one side zero, factor, and set each factor to zero. If only the square remains, keep both root signs when the right side is positive. Check every candidate in the starting formula.
An output after the equals sign is the target, not an input to substitute.
The zero product rule requires the entire product to equal zero.
Keep every candidate and check each one in the original formula.
p2 + 2p = 3.Replace the requested function output h(p) with its formula.
p2 + 2p − 3 = 0.Subtracting 3 from both sides makes zero on one side. That lets a factored product use the zero product rule to find every input.
The pairs with product −3 are 1 and −3, with sum −2, and −1 and 3, with sum 2. Choose −1 and 3.These are the constant product and middle-term sum required for factoring.
p + 3 = 0 or p − 1 = 0.The zero product rule says at least one factor must equal zero.
p + 3 − 3 = 0 − 3, so p = −3.Subtracting 3 from both sides solves the first smaller equation.
p − 1 + 1 = 0 + 1, so p = 1.Adding 1 to both sides solves the second smaller equation.
h(−3) = (−3)2 + 2(−3) = 9 − 6 = 3, and h(1) = 12 + 2(1) = 1 + 2 = 3.Each candidate must produce the required output in the original formula.
Answer
p = −3
p = 1
Check The curve meets the horizontal level line at height 3 at p = −3 and p = 1. Those graph readings agree with both substitutions in the formula.
Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.
Rung 1Factoring rung 1: pull out a common factor
Factor 2x + 6. This asks you to write the expression as a multiplication without changing its value.
Both pieces share the factor 2.
What it asks. Rewrite the same sum as multiplied factors. Do not change its value.
Plan. Identify a multiplier in every term. Write it outside parentheses and the remaining pieces inside. Distribute back to check.
Write both terms as products with the same shared number.
Each inside term is what remains after its shared multiplier is taken outside.
2x + 6 = 2 × x + 2 × 3.Both terms contain the factor 2, since 6 = 2 × 3.
2x + 6 = 2(x + 3).Reverse distribution by taking the shared factor 2 outside the parentheses.
Answer
2x + 6 = 2(x + 3)
Check Expand the answer: 2(x + 3) = 2x + 2 × 3 = 2x + 6, the original expression.
Rung 2Factoring rung 2: two positive numbers
Factor x2 + 3x + 2. Find two parentheses that multiply to the original expression.
The middle pieces 2x and x add to 3x.
What it asks. Rewrite the same sum as multiplied factors. Do not change its value.
Plan. Identify the constant and middle coefficient. List integer factor pairs of the constant, check their sums and signs, and write the pair in parentheses. Multiply the factors back to check.
The two constant numbers must multiply to the last term.
Their sum must equal the coefficient of the middle term.
Check the signs, then multiply all four term pairs back.
The desired sum is 3 and the desired product is 2.The coefficient of x and the constant come from the sum and product of the two factor numbers.
Choose 1 and 2: 1 + 2 = 3 and 1 × 2 = 2.This pair matches both required numbers.
x2 + 3x + 2 = (x + 1)(x + 2).The two constants are the numbers 1 and 2.
Answer
x2 + 3x + 2 = (x + 1)(x + 2)
Check Multiply all four pairs: (x + 1)(x + 2) = x2 + 2x + x + 2 = x2 + 3x + 2.
Rung 3Factoring rung 3: opposite signs
Factor x2 − x − 6. A missing written coefficient before x means the coefficient is −1.
2 + (−3) = −1
2 × (−3) = −6
(x + 2)(x − 3) = x2 − x − 6
Match both the sum and the product, including their signs.
What it asks. Rewrite the same sum as multiplied factors. Do not change its value.
Plan. Identify the constant and middle coefficient. List integer factor pairs of the constant, check their sums and signs, and write the pair in parentheses. Multiply the factors back to check.
The two constant numbers must multiply to the last term.
Their sum must equal the coefficient of the middle term.
Check the signs, then multiply all four term pairs back.
The desired sum is −1 and the desired product is −6.Read the middle coefficient and constant with their signs.
A negative product needs opposite signs. Choose 2 and −3.2 × (−3) = −6 and 2 + (−3) = −1.
x2 − x − 6 = (x + 2)(x − 3).Adding −3 is written as subtracting 3.
Answer
x2 − x − 6 = (x + 2)(x − 3)
Check Expand: (x + 2)(x − 3) = x2 − 3x + 2x − 6 = x2 − x − 6. The negative middle coefficient is preserved.
Rung 4Factoring rung 4: the quadratic
Factor x2 + 2x − 3. This is the same pattern used when solving h(p) = 3.
The two middle products −x and 3x combine to 2x.
What it asks. Rewrite the same sum as multiplied factors. Do not change its value.
Plan. Identify the constant and middle coefficient. List integer factor pairs of the constant, check their sums and signs, and write the pair in parentheses. Multiply the factors back to check.
The two constant numbers must multiply to the last term.
Their sum must equal the coefficient of the middle term.
Check the signs, then multiply all four term pairs back.
The desired sum is 2 and the desired product is −3.The coefficient of x is 2 and the constant is −3.
Choose 3 and −1, since 3 + (−1) = 2 and 3 × (−1) = −3.A correct pair must satisfy both conditions.
x2 + 2x − 3 = (x + 3)(x − 1).Place the two numbers into the two factors.
Answer
x2 + 2x − 3 = (x + 3)(x − 1)
Check Expand: (x + 3)(x − 1) = x2 − x + 3x − 3 = x2 + 2x − 3. Replacing x by p gives p2 + 2p − 3 = (p + 3)(p − 1).
Rung 5Factoring rung 5: two negative numbers
Factor x2 − 9x + 20. Find two parentheses that multiply back to this sum.
Both negative constants give a positive last product and a negative middle sum.
What it asks. Rewrite the same sum as multiplied factors. Do not change its value.
Plan. Identify the constant and middle coefficient. List integer factor pairs of the constant, check their sums and signs, and write the pair in parentheses. Multiply the factors back to check.
The two constant numbers must multiply to the last term.
Their sum must equal the coefficient of the middle term.
Check the signs, then multiply all four term pairs back.
The required sum is −9 and product is 20.The middle coefficient and constant determine both targets.
The positive pairs of 20 are 1 and 20, 2 and 10, 4 and 5. Their sums are 21, 12, and 9.Listing pairs prevents guessing a pair that matches only the product.
Use −4 and −5: their sum is −9 and their product is 20.A positive product and negative sum require both numbers to be negative.
x2 − 9x + 20 = (x − 4)(x − 5).The constants are the pair matching both requirements.
Let r(x) = 3x − 7. Solve r(x) = 8. Find the input that produces output 8.
After removing the subtraction, divide the equal totals into three groups.
What it asks. The output is given. Find every allowed input that produces it.
Plan. Replace the function call by its formula. Set that formula equal to the requested output to turn the backward question into an equation. Undo the added or subtracted constant on both sides, then divide both sides by the nonzero input coefficient. Check the recovered input in the original formula.
The value after the equals sign is the requested output.
Use the same operation on both sides to preserve equality.
Substitution in the starting formula must return the target output.
3x − 7 = 8.Set the formula equal to the stated output.
3x = 15.Add 7 to both sides.
x = 5.Divide both sides by 3.
Answer
x = 5
Check r(5) = 15 − 7 = 8, so the answer matches the requested output.
Rung 7Solving rung 3: a square has two input signs
For s(x) = x2, solve s(x) = 9. Find every input whose square is 9.
The two opposite inputs reach the same height 9.
What it asks. The output is given. Find every allowed input that produces it.
Plan. Set the formula equal to the requested output. If there is a middle term, subtract the output to make one side zero, factor, and set each factor to zero. If only the square remains, keep both root signs when the right side is positive. Check every candidate in the starting formula.
An output after the equals sign is the target, not an input to substitute.
The zero product rule requires the entire product to equal zero.
Keep every candidate and check each one in the original formula.
x2 = 9.Replace the function output with its formula.
9 = 3.3 × 3 = 9, so the nonnegative square root is 3.
x = 3 or x = −3.Both 3 × 3 and (−3)(−3) equal 9; squaring removes the input's sign.
Answer
x = −3
x = 3
Check s(−3) = 9 and s(3) = 9. A negative input is permitted here because the domain of the square recipe includes all real numbers.
Rung 8Solving rung 4: factor after moving the output
For h(p) = p2 + 2p, solve h(p) = 3. Find all inputs giving output 3.
Keep both checked inputs in the answer.
What it asks. The output is given. Find every allowed input that produces it.
Plan. Set the formula equal to the requested output. If there is a middle term, subtract the output to make one side zero, factor, and set each factor to zero. If only the square remains, keep both root signs when the right side is positive. Check every candidate in the starting formula.
An output after the equals sign is the target, not an input to substitute.
The zero product rule requires the entire product to equal zero.
Keep every candidate and check each one in the original formula.
p2 + 2p = 3, so p2 + 2p − 3 = 0.Set the formula equal to the output, then subtract 3 from both sides.
p2 + 2p − 3 = (p + 3)(p − 1).3 and −1 multiply to −3 and add to 2.
p + 3 = 0 or p − 1 = 0.The factors have product zero, so at least one factor is zero.
p = −3 or p = 1.Subtract 3 in the first equation and add 1 in the second.
