Polynomial Arithmetic and Factoring

Part 1 · Hook

Why this matters

An engineer who knows the outputs of a curve's formula still needs its zeros: where a beam's bending is zero, where a profit breaks even. For the curves in this unit, the zeros come from writing the formula as a product. That is factoring, and it is the algebra Topic 1.4 builds on.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Solve $x^2 - 5x + 6 = 0$ .

$x = 2$ or $x = 3$
$x = -2$ or $x = -3$
$x = 5$ or $x = 6$
$x = 1$ or $x = 6$

Show the answer

$(x - 2)(x - 3) = 0$ .

Correct: $x = 2$ or $x = 3$ :
$x = -2$ or $x = -3$ :
$x = 5$ or $x = 6$ :
$x = 1$ or $x = 6$ :

2. What is $(x + 3)^2$ multiplied out?

$x^2 + 6x + 9$
$x^2 + 9$
$x^2 + 3x + 9$
$x^2 + 6x + 6$

Show the answer

$(x + 3)(x + 3)$ $= x^2 + 3x + 3x + 9$ .

Correct: $x^2 + 6x + 9$ :
$x^2 + 9$ :
$x^2 + 3x + 9$ :
$x^2 + 6x + 6$ :

Part 4 · See it

See it first

The graph of y = x³ − 9x = x(x − 3)(x + 3). It crosses the x-axis at x = −3, 0 and 3, one crossing for each factor: x + 3 = 0, x = 0 and x − 3 = 0. — $y$ $= x^3 - 9x$ $= x(x - 3)(x + 3)$ crosses the $x$ -axis once for each factor: at $x = -3$ , $0$ and $3$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Every term shares a factortake it out first: $x^3 - 9x = x(x^2 - 9)$
A factor matches a pattern ( $a^2 - b^2$ , $a^3 - b^3$ , $a^3 + b^3$ )use the pattern: $x^2 - 9 = (x - 3)(x + 3)$
Four terms pair up with a shared binomialfactor by grouping: $x^2(x + 2) - 4(x + 2)$ $= (x^2 - 4)(x + 2)$
A polynomial is written as a product equal to $0$ each factor equal to $0$ gives a solution, and $p(a) = 0$ means $x - a$ is a factor

Part 6 · Key ideas

Key ideas

A polynomial is a sum of terms $ax^n$ with whole-number powers, like $x^3 + x^2 - 7x + 2$ .
Multiply by distributing every term of one factor over every term of the other, then combine like terms.
Factor in order: common factor, then patterns (difference of squares, sum or difference of cubes), then grouping.

Part 7 · Misconception

A common mistake

The wrong idea: To solve $x^3 = 9x$ , divide both sides by $x$ to get $x^2 = 9$ .

What actually happens: Dividing by $x$ assumes $x \ne 0$ and loses the solution $x = 0$ . Move everything to one side and factor: $x(x - 3)(x + 3) = 0$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

1. Multiply out $(x - 2)(x^2 + 3x - 1)$ and combine like terms.

Type an expression.

Show the answer

$x(x^2 + 3x - 1) - 2(x^2 + 3x - 1)$ $= x^3$ $+ 3x^2 - x - 2x^2 - 6x$ $+ 2$ $= x^3$ $+ x^2 - 7x$ $+ 2$ .

Answer: $x^{3} + x^{2} - 7x + 2$

2. Which is $x^3 - 9x$ factored completely?

$x(x^2 - 9)$
$x(x - 3)(x + 3)$
$(x - 3)(x + 3)$
$x(x - 3)(x - 3)$

Show the answer

Common factor first: $x^3 - 9x = x(x^2 - 9)$ . Then $x^2 - 9 = (x - 3)(x + 3)$ . So $x^3 - 9x$ $= x(x - 3)(x + 3)$ .

$x(x^2 - 9)$ : $x^2 - 9$ is a difference of squares and still splits: $(x - 3)(x + 3)$ .
Correct: $x(x - 3)(x + 3)$ : Right: take out the common $x$ , then factor $x^2 - 9$ as a difference of squares.
$(x - 3)(x + 3)$ : This multiplies to $x^2 - 9$ ; the common factor $x$ was dropped.
$x(x - 3)(x - 3)$ : $(x - 3)(x - 3)$ $= x^2 - 6x + 9$ , not $x^2 - 9$ .

3. Factor $x^3 + 2x^2 - 4x - 8$ by grouping.

$(x + 2)(x^2 + 4)$
$(x - 2)(x + 2)$
$(x + 2)^2(x + 2)$
$(x - 2)(x + 2)^2$

Show the answer

Group: $x^2(x + 2) - 4(x + 2)$ . Factor out $(x + 2)$ : $(x^2 - 4)(x + 2)$ . Then $x^2 - 4 = (x - 2)(x + 2)$ , so the result is $(x - 2)(x + 2)^2$ .

$(x + 2)(x^2 + 4)$ : The second group is $-4(x + 2)$ , so the other factor is $x^2 - 4$ , not $x^2 + 4$ .
$(x - 2)(x + 2)$ : This has only two linear factors; it multiplies to $x^2 - 4$ , not a cubic.
$(x + 2)^2(x + 2)$ : That is $(x + 2)^3$ ; check: its constant term is $8$ , not $-8$ .
Correct: $(x - 2)(x + 2)^2$ : Right: $x^2(x + 2) - 4(x + 2)$ $= (x^2 - 4)(x + 2)$ $= (x - 2)(x + 2)(x + 2)$ .

4. For $p(x) = x^3 - 4x^2 + x + 6$ , $p(2) = 0$ . Which must be a factor of $p(x)$ ?

$x + 2$
$x - 6$
$x - 2$
$x$

Show the answer

If $p(a) = 0$ , then $x - a$ is a factor. Here $p(2)$ $= 8 - 16 + 2 + 6$ $= 0$ , so $x - 2$ is a factor; in fact $p(x)$ $= (x - 2)(x - 3)(x + 1)$ .

$x + 2$ : $x + 2$ is $0$ at $x = -2$ ; the factor for the zero $x = 2$ is $x - 2$ .
$x - 6$ : $6$ is the constant term; $p(6) = 84 \ne 0$ .
Correct: $x - 2$ : Right: an input $a$ with $p(a) = 0$ gives the factor $x - a$ .
$x$ : $p(0) = 6 \ne 0$ , so $x$ is not a factor.

5. Which is $x^4 - 5x^2 + 4$ factored completely?

$(x^2 - 1)(x^2 - 4)$
$(x - 1)(x + 1)(x - 2)(x + 2)$
$(x - 1)^2(x - 2)^2$
$(x^2 + 1)(x^2 + 4)$

Show the answer

Let $u = x^2$ : $u^2 - 5u + 4$ $= (u - 1)(u - 4)$ . So $x^4 - 5x^2$ $+ 4$ $= (x^2 - 1)(x^2 - 4)$ $= (x - 1)(x + 1)(x - 2)(x + 2)$ .

$(x^2 - 1)(x^2 - 4)$ : Correct as a first step, but each factor is a difference of squares and splits again.
Correct: $(x - 1)(x + 1)(x - 2)(x + 2)$ : Right: treat it as a quadratic in $x^2$ , then split each difference of squares.
$(x - 1)^2(x - 2)^2$ : This has $x^3$ and $x$ terms when multiplied out; $x^4 - 5x^2 + 4$ has none.
$(x^2 + 1)(x^2 + 4)$ : This multiplies to $x^4 + 5x^2 + 4$ : the sign of the middle term is wrong.

6. What are the solutions of $x^3 - 9x = 0$ ?

$x = 3$ and $x = -3$
$x = 9$
$x = 0$ , $x = 3$ and $x = -3$
$x = 0$ , $x = 9$ and $x = -9$

Show the answer

Factor: $x(x - 3)(x + 3) = 0$ . A product is $0$ when a factor is $0$ : $x = 0$ , $3$ or $-3$ . Never divide by $x$ ; it hides $x = 0$ .

$x = 3$ and $x = -3$ : Dividing both sides by $x$ loses the solution $x = 0$ . Factor instead.
$x = 9$ : $x^3 = 9x$ does not mean $x = 9$ ; $9^3 - 81 \ne 0$ .
Correct: $x = 0$ , $x = 3$ and $x = -3$ : Right: $x(x - 3)(x + 3) = 0$ , and each factor gives a solution.
$x = 0$ , $x = 9$ and $x = -9$ : $x^2 = 9$ gives $x = \pm 3$ , not $\pm 9$ .

Part 9 · Summary

Summary

Multiply polynomials by distributing every term and combining like terms. Factor by taking out a common factor first, then using a difference of squares

a^2 - b^2 = (a - b)(a + b)

, a sum or difference of cubes, or grouping. A product equals

0

when a factor does, and

p(a) = 0

means

x - a

is a factor.

Part 10 · Up next

What comes next

Part 11 · Connections