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McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
Study Guide and Review
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Exercise 64 Page 83

Divide the area by the length to find the width.

w=3x+7 units

Practice makes perfect

To find the missing side length of this rectangle we will need to factor. The area is a trinomial, 6x^2+11x-7. We can solve by factoring the trinomial where a=6, b=11, and c = -7. We need to find two factors of ac=-42 that have a sum of b.

Factors of -42 Sum of factors
-1 and 42 41
-2 and 21 19
-3 and 14 11
Since -3 and 14 are the values we seek, let's split our linear term into 11x=-3x + 14x then factor.

6x^2+11x-7
â–¼
Factor
Substitute

11x= -3x + 14x

6x^2 -3x + 14x-7
FactorOut

Factor out 3x

3x(2x-1) + 14x-7
FactorOut

Factor out 7

3x(2x-1) + 7(2x-1)
FactorOut

Factor out 2x-1

(3x+7)(2x-1)



Since we already have the length l=2x-1, the other factor must be the width. w=3x+7

Alternative Solution

 Using 2x-1
The area is a trinomial, 6x^2+11x-7 and width is a binomial, 2x-1. Essentially, we need to find a missing factor, ax+b to multiply by 2x-1 to get 6x^2+11x-7. & 6x^2+ 11x+(-7) &= (2x-1)(ax + b) &= 2ax^2+( 2b-a)x+(-b) In this scenario, 6 = 2a and - 7= - b. Let's solve for a and b.

lc6 =2a & (I) - 7= - b & (II)
DivEqn

(I): .LHS /2.=.RHS /2.

l3 = a - 7= - b
DivEqn

(II): .LHS /(- 1).=.RHS /(- 1).

l3 = a 7= b

(I), (II):Rearrange equation

la=3 b=7

Now, we can say the second factor representing the width of the rectangle is 3x+7.

Subchapter links expand_more
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Adding and Subtracting Polynomials 12 (Page 81)
Adding and Subtracting Polynomials 13 (Page 81)
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arrow_right 7 Solving ax^2+bx+c=0 p.83
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