Question

Let $f(x) = x^{10}-8x^8-8x^3+12x^2-5x-5$. Without using long division (which would be horribly nasty!), find the remainder when $f(x)$ is divided by $x^2-1$.

Answer 1

Answer:

-13x + 10

Step-by-step explanation:

We have the polynomial $f(x) = x^{10}-8x^{8}-8x^{3}+12x^{2}-5x+5$.

It is required to find the remainder when f(x) is divided by $g(x)=x^{2}-1$.

Using the division algorithm, we get there exists polynomials q(x) and r(x) such that f(x) = g(x) q(x) + r(x), where q(x) is the quotient and r(x) is the remainder with 0 < degree of r(x) < degree of g(x) = 2.

This gives r(x) = ax + b, where a and b are integers.

Now, $f(x) = (x-1) \times (x+1) \times q(x)$

So, f(1) = r(1) and f(-1) = r(-1)

i.e. -3 = a+b and 23 = -a+b

On solving we get, a = -13 and b = 10.

So, r(x) = -13x + 10.

Hence the remainder without using long division is -13x+10.

Answer 2

Solution:

$f(x) = x^{10}-8x^8-8x^3+12x^2-5x-5$

We have to find the remainder when f(x) is divided by $x^2-1.$

x²-1=0

x=$\pm1$

$f(1)=1^{10}-8(1)^8-8(1)^3+12(1)^2-5(1)-5=1-8-8+12-5-5=1-16+12-10=13-26=-13\\\\f(-1)=(-1)^{10}-8(-1)^8-8(-1)^3+12(-1)^2-5(-1)-5=1-8+8+12+5-5=1+12=13$

So, remainder is 13 and -13.