Find the values of R, P, and Q that make the polynomial

Ответ:

benjaminmccutch

04.02.2022

Given:

$\left(12x^5+Rx^4+Px^2-5\right)+\left(-6x^5-4x^4+13x+Q\right)=6x^5+3x^4+9x^2+13x+7$

To find:

The values of R, P and Q.

Solution:

We have,

$\left(12x^5+Rx^4+Px^2-5\right)+\left(-6x^5-4x^4+13x+Q\right)=6x^5+3x^4+9x^2+13x+7$

On combining like terms, we get

(12x^5-6x^5)+(Rx^4-4x^4)+Px^2+13x+(Q-5)=6x^5+3x^4+9x^2+13x+7

6x^5+(R-4)x^4+Px^2+13x+(Q-5)=6x^5+3x^4+9x^2+13x+7

On comparing the coefficients of x^4 , we get

R-4=3

R=3+4

R=7

On comparing the coefficients of x^2 , we get

P=9

on comparing the constants, we get

Q-5=7

Q=7+5

Q=12

Therefore, R=7,P=9 and Q=12 .

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