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All courses Math III · F-IF.8 27 of 55
Rewrite functions in equivalent forms to reveal and explain useful properties.

Rewrite a polynomial in factored form to reveal its zeros

Problem
What factored form would reveal the zeros of x²-5x+6?
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Hint

Factor the quadratic into two binomials. Look for two numbers that multiply to 6 and add to -5.

For x²+bx+c, look for two numbers whose product is c and whose sum is b. Those numbers become the constants in the binomial factors.

Solution walkthrough

01

Set up the factoring target

\[\text{product}~6,~\text{sum}~-5\]

To factor x²-5x+6, look for two numbers whose product is 6 and whose sum is -5.

02

Find the factor pair

\[(-2)(-3)=6~\text{and}~-2+(-3)=-5\]

The numbers -2 and -3 multiply to 6 and add to -5, so they give the correct binomial factors.

03

Write the factored form

\[x^2-5x+6~=~(x-2)(x-3)\]

Using those two numbers in the binomials rewrites the quadratic as a product without changing its value.

04

State the requested form

\[\text{Factored}~\text{form}:~(x-2)(x-3)\]

This factored form reveals the zeros because each factor can be set equal to 0: x-2=0 gives 2 and x-3=0 gives 3.

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Another way

  1. You can check by expanding: (x-2)(x-3)=x²-5x+6, so the factorization matches the original expression.

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Common mistake

A common mistake is using (x+2)(x+3). That expands to x²+5x+6, which has the wrong middle term sign.