Question

Find the values of R, P, and Q that make the polynomial addition problem true. Write your answer in this order: R, P, Q. \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 (12x 5 +Rx 4 +Px 2 −5)+(−6x 5 −4x 4 +13x+Q)=6x 5 +3x 4 +9x 2 +13x+7

Answer 1

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`.