Question

if a polynomial is divided by x-5, the quotient is 2x^2 +8x-9 and the remainder is 3 what is the original polynomial?

Answer 1

Answer: 2x^3 - 2x^2 - 49x + 48

Step-by-step explanation:

Let a be the original polynomial

Where a/(x-5) = 2x^2 + 8x - 9 (remainder 3)

a = (2x^2 + 8x - 9)(x-5) + 3
a = 2x^3 - 10x^2 + 8x^2 - 40x - 9x + 45+3
a = 2x^3 - 2x^2 - 49x + 48

Answer 2

`P(x)=2x^3-2x^2-49x+48`

Step-by-step explanation:

Let the original polynomial be
`P(x)`.

We know that when it is divided by
`(x-5)`, the quotient is
`2x^2+8x-9` and we get a remainder of 3.

Therefore, this means that:

`(P(x))/(x-5)=2x^2+8x-9+(3)/(x-5)`

Remember what it means when we have a remainder. Say we have 13 divided by 3. Our quotient will be 4 R1, or 4 1 over 3. We put the remainder over the divisor. This is the same thing for polynomials.

So, to find our original polynomial, multiply both sides by
`(x-5)`:

`(x-5)(P(x))/(x-5)=(x-5)(2x^2+8x-9+(3)/(x-5))`

The left side will cancel. Distribute the right:

`P(x)=2x^2(x-5)+8x(x-5)-9(x-5)+(3)/((x-5))(x-5)`

Distribute:

`P(x)=(2x^3-10x^2)+(8x^2-40x)+(-9x+45)+(3)`

Combine like terms:

`P(x)=(2x^3)+(-10x^2+8x^2)+(-40x-9x)+(45+3)`

Evaluate:

`P(x)=2x^3-2x^2-49x+48`

And we're done!