# Need help solving this question (Remainder Theorem)

I know, it's probably an easy question for most of you people, but I really need help and if any one could explain step by step how to do this, that'd be great,

Question One:

The expression x² + bx + a leaves the same remainder when divided by x + 2 or by x -a, where a ≠ -2. Show that a + b = 2

A similar question to the one above:

When 2x³ - 4x² - 5x - 2 is divided by (x - 1)(x + 2), the remainder is ax + b. The result may be expressed as the identity:

2x³ - 4x² - 5x - 2 ≡ (x - 1)(x + 2)Q(x) + ax + b , Where Q(x) is the quotient.

(a) State the degree of Q(x) Can someone please explain what do they mean by degree? (b) By substituting suitable values of x, find a and b

First question
$f(x)=x^2+bx+a$ by remainder theorem and the condition given in the problem we have $f(-2)=f(a)$, that means $4-2b+a=a^2+ab+a$ which after factorized is equivalent to $(a+2)(a+b-2)=0$. Because $a\neq-2$, $a+b=2$

Degree is the power of higest power term in polynomial. For example polynomial $x^{100}+8x^{37}+1$ has degree 100, constant has degree 0, etc.

By comparsion it's clear $Q(x)$ has degree 1.

Plug in $x=1$ to the identity you get $-9=a+b$
Plug in $x=-2$ to the identity you get $-24=-2a+b$
Solve a,b you get $a=5, b=-14$

• Thanks for the explanation!:) – Talha Tanveer Sep 15 '14 at 16:51

For the question one :

Let $f(x)=x^2+bx+a$. Since $f(-2)=f(a)$ by the polynomial remainder theorem with $a+2\not =0$, we have $$(-2)^2-2b+a=a^2+ab+a\Rightarrow (a+2)(a+b-2)=0\Rightarrow a+b=2.$$

For the question two :

The degree of a polynomial is the highest degree of any term in the polynomial. So, for example, the degree of the following polynomial is $3$ : $$2x^3-4x^2-5x-2.$$ Here, note that the degree of $(x-1)(x+2)$ is $2$ and that the degree of $ax+b$ is smaller than or equal to $1$.

Are a and b integers? Suppose that. A polynomial $P(x)$ when divided by $x-b$, leaves remainder P(b). To see this, write $P(x) = (x-b)Q(x) + r$, where r is a constant (Why can this be written?) . Then, P(b) = r. So, in your problem, $P(-2) = P(a)$, where P is the given quadratic. Hence, $4 - 2b + a = a^2 +ab + a$ so that $a^2 - 4 + 2b + ab = 0$ so that $(a + 2)(a - 2 + b) = 0$. Now, you are done!