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

1 Answer
Feb 29, 2016

Answer:

This is a perfect square trinomial, because the first and last terms are perfect squares #(sqrt(x^2) = x and sqrt(16) = 4)#

Explanation:

Method 1:

#(x - 4)(x - 4)#

#(x - 4)^2#

You can also double check by making sure term b (the middle term) satisfies the equation #b = 2ac# only once you have factored, or when you have taken the square root of the first and last term. We check: #8 = 2(x)(4)#. 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 #ax^2 + bx + 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 + 8x + 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 #(x - 4)^2#

Practice exercises:

  1. Factor the following trinomials using method 1

a) #x^2 + 10x + 25#

b) #16x^2 - 56x + 49#

2 . Factor the following trinomials using method 2

a) #x^2 - 22x + 121#

b) #x^2 + 5x + 6#

c) #x^2 - 8x - 33#

d) #x^2 - 14x + 45#

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

a) #4x^2 + mx + 64#

b) #25x^2 - 40x + m#

Good luck!