McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
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Exercise 77 Page 84

To factor a perfect square trinomial, the first and last terms have to be perfect squares.

(2-7a)^2

Practice makes perfect
We want to factor a perfect square trinomial. 4-28a+49a^2

How do we know that the expression is a perfect square trinomial? Well, let's ask a few questions.

Is the first term a perfect square? 4= 2^2 âś“
Is the last term a perfect square? 49a^2= 7^2 a^2 âś“
Is the middle term twice the product of 14 and a? 28a=2* 14* a âś“
As we can see, the answer to the three questions above is yes! Therefore, we can write the trinomial as the square of a binomial. Note there is a subtraction sign in the middle. 4-28a+49a^2 ⇔ ( 2- 7a)^2

Checking Our Answer

Check your answer âś“
Let's un-factor our answer and compare it with the given expression.
(2-7a)^2
2^2-2(2)(7a)+(7a)^2
2^2-28a+(7a)^2
2^2-28a+7^2a^2
4-28a+49a^2
After expanding and simplifying, the result is the same as the given expression. Therefore, we can be sure our solution is correct!