Chapter 5.9, Problem 40WE

To determine

To calculate: the prime of $2{\left(x-y\right)}^2-9\left(x-y\right)z-5z^2$

Answer to Problem 40WE

  $2{\left(x-y\right)}^2-9\left(x-y\right)z-5z^2=\left(2x-2y+z\right)\left(x-y-5z\right)$

Explanation of Solution

Given information: The given equation is $2{\left(x-y\right)}^2-9\left(x-y\right)z-5z^2$

Calculation: Here we use splitting the middle term formula, here if you want you can assume $\left(x-y\right)=t$ for simplicity then this is quadratic equation

We have to split the middle term in two parts so that the sum of both terms is equal to $-9$ and product of the both terms is $2\left(-5\right)=-10$ .

First we find all the possible factors of $-10$ ,

$-\left(1\right)\left(10\right)$ , $-\left(2\right)\left(5\right)$ .

we try all the possibility, if any of them satisfy then we proceed with it else write prime

$-10+1=-9$ ,

so required factors are $-10$ and $1$

Now we can re-write the given equation

$2{\left(x-y\right)}^2-9\left(x-y\right)z-5z^2=2{\left(x-y\right)}^2-10\left(x-y\right)z+\left(x-y\right)z-5z^2$

  $=\left(x-y\right)\left(2x-2y+z\right)-5z\left(2x-2y+z\right)$

Taking $\left(2x-2y+z\right)$ common from both the terms, we get

$=\left(2x-2y+z\right)\left(x-y-5z\right)$

  $2{\left(x-y\right)}^2-9\left(x-y\right)z-5z^2=\left(2x-2y+z\right)\left(x-y-5z\right)$

Conclusion: The factors of the polynomial are $\left(2x-2y+z\right)$ and $\left(x-y-5z\right)$ , hence

  $2{\left(x-y\right)}^2-9\left(x-y\right)z-5z^2=\left(2x-2y+z\right)\left(x-y-5z\right)$