# Polynomial Division Question?

I'm given a question such as this. Use the remainder theorem to find the value of "k" in each polynomial.

I'm then given a question such as (2x^3 + 3x^2 + kx -3) / (2x+5) Remainder = 2

Can somebody please kindly explain to me what I should be do?

Note - I'm an adult who is self-studying.

I do have the book "The Humongous Book of Algebra Problems" and the author does have two of these questions. Page 254, & 255, but he doesn't explain any further. I've tried converting the (2x+5) into x=-5/2) or 2.5 and tried calculating like that I still don't get the same answers as I think I should be getting.

### 4 Answers

- PuzzlingLv 72 months agoFavorite Answer
In general, if you have a polynomial p(x) and a linear factor of the form x - a, when you divide you get a quotient (q(x)) and a remainder r.

The Polynomial Remainder Theorem gives us a shortcut to finding out the remainder, by just computing f(a) = r.

So as a shortcut, you can just figure out 'a', calculate f(a) and then solve for k.

Example:

(2x^3 + 3x^2 + kx -3) / (2x+5)

Looking at the linear factor:

2x + 5 = 0

2x = -5

x = -5/2

Now plug that into the original function and set it to the remainder 2:

2(-5/2)^3 + 3(-5/2)^2 + k(-5/2) - 3 = 2

2(-125/8) + 3(25/4) + (-5/2)k - 3 = 2

-125/4 + 75/4 + (-5/2)k = 5

(-5/2)k = 5 + 50/4

(-5/2)k = 5 + 25/2

(-5/2)k = 10/2 + 25/2

(-5/2)k = 35/2

k = (35/2) * (-2/5)

k = -7

Read more in the link below.

- Ian HLv 72 months ago
Explaining the ideas behind questions like this.

We can easily invent a polynomial that has remainder 2 like this.

P(x) = (x – 3)(x – 5) + 2 = x^2 – 8x + 19 ..........(1)

P/(x – 3) = (x – 5) + 2/(x – 3)

But substituting x = 3 into (1) makes the first expression zero so that P(3) = 2

That is the idea behind the remainder theorem; as puzzling wrote, it gives

us a shortcut to finding out the remainder, by just computing f(a) = r.

Usually you get a question in this form. What is the remainder when

P(x) = 2x^3 + 3x^2 – 7x – 3 ? Long division has shown us that

P(x) = (x^2 – x – 1)*(2x + 5) + 2,

so we can see why substituting x = -5/2 just would leave remainder = 2

However this question is the other way round; we are told remainder = 2

with P(x) = 2x^3 + 3x^2 + kx – 3 and we are asked to find k

We just do the same and substituting x = -5/2 eventually leads to

-(5/2)k – 31/2 = 2 which finally produces k = -7

- Wayne DeguManLv 72 months ago
Let's say we are dividing a polynomial P(x) by (x - a).

so, P(x)/(x - a) = Q(x) + R/(x - a)

=> P(x) = (x - a)Q(x) + R

Hence, P(a) = R

Now, if you are dividing by 2x + 5, then we need the value of x that makes this zero.

i.e. 2x + 5 = 0

so, x = -5/2

Then, substituting this into our polynomial we get:

2(-5/2)³ + 3(-5/2)² + k(-5/2) - 3 = remainder

so, -125/4 + 75/4 - 5k/2 - 3 = 2

Hence, -125/4 + 75/4 - 5k/2 = 5

=> (-125 + 75 - 10k)/4 = 20/4

i.e. -125 + 75 - 10k = 20

so, -50 - 10k = 20

=> 10k = -70

Then, k = -7

so, our polynomial is 2x³ + 3x² - 7x - 3

Then, (2x³ + 3x² - 7x - 3)/(2x + 5) = Q(x) + 2/(2x + 5)

so, 2x³ + 3x² - 7x - 3 = (2x + 5)Q(x) + 2

Further inspection will tell us that Q(x) must be a quadratic, i.e. ax² + bx + c

some algebraic work will give:

2x³ + 3x² - 7x - 3 = (2x + 5)(x² - x - 1) + 2

Checking with long division yields:

...........x² - x - 1

..........___________________

2x + 5) 2x³ + 3x² - 7x - 3

............2x³ + 5x²

....................-2x² - 7x

....................-2x² - 5x

............................-2x - 3

............................-2x - 5

....................................2...remainder

:)>

- DixonLv 72 months ago
The basic approach with this sort of question is not to worry about the fact that there is another unknown in the polynomial and just do what you would normally do anyway and treat the unknown as if it is some constant number - which it is.

So just do the long division, treating k as a number when it turns up. You will get a quotient and a remainder in terms of k and and x. Then put the remainder = 2. This suggests that the remainder you get won't actually have any x terms in it and you can solve for k.