Remainder theorem

**Ilsa** · April 10th 2011, 08:56 AM

The expression $f(x) = ax^3 - ( a + 3b )x^2 + 2bx + c$ is excatly divisible by $x^2 - 2x$ . When $f(x)$ is divided by $x - 1$ , the remainder is 8 more than when it is divided by $x + 1$ . Factorise $f(x)$ completely.

**topsquark** · April 10th 2011, 09:04 AM

Originally Posted by Ilsa

The expression $f(x) = ax^3 - ( a + 3b )x^2 + 2bx + c$ is excatly divisible by $x^2 - 2x$ . When $f(x)$ is divided by $x - 1$ , the remainder is 8 more than when it is divided by $x + 1$ . Factorise $f(x)$ completely.

Well, f(x) is divisible by $x^2 - 2x = x(x - 2)$ so we know f(x) is divisible by x. Thus c = 0 immediately. Consider then
$f(x) = ax^3 - (a + 3b)x^2 + 2bx$

From the original statement (divisible by x(x - 2)), we get
$f(2) = a(2)^3 - (a + 3b)(2)^2 + 2b(2) = 0$

Call
$f(1) = a(1)^3 - (a + 3b)(1)^2 + 2b(1) = d$

Then we know that
$f(-1) = a(-1)^3 - (a + 3b)(-1)^2 + 2b(-1) = d - 8$

Three equations, three unknowns.

-Dan

**Ilsa** · April 10th 2011, 09:26 AM

Originally Posted by topsquark

Well, f(x) is divisible by $x^2 - 2x = x(x - 2)$ so we know f(x) is divisible by x. Thus c = 0 immediately. Consider then
$f(x) = ax^3 - (a + 3b)x^2 + 2bx$

From the original statement (divisible by x(x - 2)), we get
$f(2) = a(2)^3 - (a + 3b)(2)^2 + 2b(2) = 0$

Call
$f(1) = a(1)^3 - (a + 3b)(1)^2 + 2b(1) = d$

Then we know that
$f(-1) = a(-1)^3 - (a + 3b)(-1)^2 + 2b(-1) = d - 8$

Three equations, three unknowns.

-Dan

If f(x) is divisible by x, how is c=0?

**topsquark** · April 10th 2011, 01:48 PM

$f(x) = ax^3 - ( a + 3b )x^2 + 2bx + c$ is excatly divisible by $x^2 - 2x$ .

If x divides f(x) then
$\displaystyle \frac{f(x)}{x}$ leaves no remainder. Thus

$\displaystyle \frac{ax^3 - (a + 3b)x^2 + 2bx + c}{x} = ax^2 - (a + 3b)x + 2b + \frac{c}{x}$

This gives a remainder of c, but f(x) is divisible by x, thus c = 0.

You can do this quicker by using the fact that if f(x) is divisible by x - r then f(r) = 0. In this case r = 0, so f(0) = 0.

-Dan

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