Answer
Chapter 6 - Section 6.2 - Exercise Set: 69 (Answer)
$x^3y^2 + x^2y - 20x$ = $x(xy + 5)(xy - 4)$
Work Step by Step
Chapter 6 - Section 6.2 - Exercise Set: 69 (Solution)
Factorize : $x^3y^2 + x^2y - 20x$
First step : Take out the GCF of $x^3y^2$, $x^2y$ and $20x$ which is $x$.
$x^3y^2 + x^2y - 20x$ = $x(x^2y^2 + xy - 20)$
Take $(x^2y^2 + xy - 20)$ to be $(xy + \triangle)(xy + \square)$
For this, we have to look for two numbers whose product is -20 and whose sum is 1.
Factors of -20 $\Longleftrightarrow$ Sum of Factors
1,-20 $\Longleftrightarrow$ -19 (Incorrect sum)
2,-10 $\Longleftrightarrow$ -8 (Incorrect sum)
4,-5 $\Longleftrightarrow$ -1 (Incorrect sum)
5,-4 $\Longleftrightarrow$ 1 (Correct sum, so the two numbers are 5 and -4)
10,-2 $\Longleftrightarrow$ 8 (Incorrect sum)
20,-1 $\Longleftrightarrow$ 19 (Incorrect sum)
Thus, $x^3y^2 + x^2y - 20x$ = $x(xy + 5)(xy - 4)$
