Answer
Chapter 6 - Section 6.2 - Exercise Set: 67 (Answer)
$\frac{1}{2}y^2 - \frac{9}{2}y - 11$ = $\frac{1}{2}(y + 2)(y - 11)$
Work Step by Step
Chapter 6 - Section 6.2 - Exercise Set: 67 (Solution)
Factor : $\frac{1}{2}y^2 - \frac{9}{2}y - 11$
First, take out the GCF of $\frac{1}{2}$ from the polynomial
$\frac{1}{2}y^2 - \frac{9}{2}y - 11$ = $\frac{1}{2}(y^2 - 9y - 22)$
Let $(y^2 - 9y - 22)$ = $(y + \triangle)(y + \square)$
Next, to look for two numbers whose product is -22 and whose sum is -9.
Factors of -22 $\Longleftrightarrow$ Sum of Factors
1,-22 $\Longleftrightarrow$ -21 (Incorrect sum)
2,-11 $\Longleftrightarrow$ -9 (Correct sum, so the two numbers are 2 and -11)
11,-2 $\Longleftrightarrow$ 9 (Incorrect sum)
22,-1 $\Longleftrightarrow$ 21 (Incorrect sum)
Thus, $(y^2 - 9y – 22)$ = $(y + 2)(y - 11)$
And, $\frac{1}{2}y^2 - \frac{9}{2}y - 11$ = $\frac{1}{2}(y + 2)(y - 11)$