How do you factor completely x^2 - 8x + 16?

1 Answer
Feb 29, 2016

This is a perfect square trinomial, because the first and last terms are perfect squares $\left(\sqrt{{x}^{2}} = x \mathmr{and} \sqrt{16} = 4\right)$

Explanation:

Method 1:

$\left(x - 4\right) \left(x - 4\right)$

${\left(x - 4\right)}^{2}$

You can also double check by making sure term b (the middle term) satisfies the equation $b = 2 a c$ only once you have factored, or when you have taken the square root of the first and last term. We check: $8 = 2 \left(x\right) \left(4\right)$. So, we have factored the trinomial properly. Also, you can check by doing FOIL (first, outside, inside and last), multiplying out.

Method 2:

Factor as a regular trinomial of the form $a {x}^{2} + b x + c , a = 1$. This method, although longer, is good to get used to because you will have to learn at one point to factor trinomials such as ${x}^{2} + 8 x + 15$, and it is the most safe and foolproof method.

To factor a trinomial of the form ax^2 + bx + c, a = 1#, you must find two numbers that multiply to c and that add to b.

We must find two numbers that multiply to +16 and add to -8. These two numbers are -4 and -4.

So, (x - 4)(x - 4). Since the parentheses repeats itself twice, we can rewrite the expression as ${\left(x - 4\right)}^{2}$

Practice exercises:

1. Factor the following trinomials using method 1

a) ${x}^{2} + 10 x + 25$

b) $16 {x}^{2} - 56 x + 49$

2 . Factor the following trinomials using method 2

a) ${x}^{2} - 22 x + 121$

b) ${x}^{2} + 5 x + 6$

c) ${x}^{2} - 8 x - 33$

d) ${x}^{2} - 14 x + 45$

3 . Find the value of $m$ that makes the following trinomials perfect square trinomials

a) $4 {x}^{2} + m x + 64$

b) $25 {x}^{2} - 40 x + m$

Good luck!