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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^2How 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 ✓
Is the last term a perfect square?	49a^2= 7^2 a^2 ✓
Is the middle term twice the product of 14 and a?	28a=2* 14* a ✓

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 ✓

Let's un-factor our answer and compare it with the given expression.

(2-7a)^2

ExpandNegPerfectSquare

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

2^2-2(2)(7a)+(7a)^2

Multiply

2^2-28a+(7a)^2

PowProdII

(a b)^m=a^m b^m

2^2-28a+7^2a^2

CalcPow

Calculate power

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!

Subchapter links

1 Adding and Subtracting Polynomials p.81

2 Multiplying a Polynomial by a Monomial p.81

3 Multiplying Polynomials p.81

4 Special Products p.82

5 Using the Distributive Property p.82

6 Solving x^2+bx+c=0 p.83

7 Solving ax^2+bx+c=0 p.83

8 Differences of Squares p.84

9 Perfect Squares p.84

10 Roots and Zeros p.85

Exercise 77 Page 84

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