# Chapter 6 - Section 6.2 - Factoring Trinomials of the Form x2+bx+c - Exercise Set: 69

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, $x^3y^2 + x^2y - 20x$ = $x(xy + 5)(xy - 4)$

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