quotient of polynomial

**rcs** · September 9th 2012, 01:47 AM

when ax^ - 4x + 4 is divided by x - 2, the remainder is 4, what is the quotient?

**kalyanram** · September 9th 2012, 02:23 AM

The exponent of x in the first term is missing but here is the general procedure.

We have $ax^\alpha - 4x + 4 = P(x)(x-2)+4$ , where $P(x)$ is a polynomial of degree $\alpha-1$
$\implies (ax^\alpha - 4x + 4) - 4 = P(x)(x-2) \implies ax^\alpha - 4x = P(x)(x-2) \implies$ 2 is a root of $ax^\alpha - 4x$ $\implies a2^\alpha = 8 \implies a = \frac{8}{2^\alpha}$

**rcs** · September 9th 2012, 04:15 AM

when ax^2 - 4x + 4 is divided by x - 2, the remainder is 4, what is the quotient? sorry this should have been the correct one...

**skeeter** · September 9th 2012, 06:01 AM

Originally Posted by rcs

when ax^2 - 4x + 4 is divided by x - 2, the remainder is 4, what is the quotient? sorry this should have been the correct one...

perform synthetic division

**rcs** · September 9th 2012, 06:03 AM

it is impossible because the numerical coefficient in first term is variable a

**skeeter** · September 9th 2012, 06:20 AM

Originally Posted by rcs

it is impossible because the numerical coefficient in first term is variable a

is that so?

Code:

2]   a ..... -4 ...... 4
.............2a ....(4a-8)
--------------------------
.....a....(2a-4) ...(4a-4)

$4a-4$ is the remainder ... set that equal to 4 and solve for $a$

... note you could also use the remainder theorem to answer this question.

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