How do you divide $\frac{x^{3} - 2 x^{2} - 3 x + 2}{x - 1}$ $(x^3-2x^2-3x+2) / (x-1)$ ?

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Answer 1 · 2016-10-01T18:36:32

$The Quotient Poly. = x^{2} - x - 4$ $"The Quotient Poly."=x^2-x-4$ .

$The Remainder= - 2$ $"The Remainder="-2$ .

We can perform the Long Division and get the Quotient and the

Remainder. But, here is another way to solve the Problem.

Suppose that, when $P (x) = x^{3} - 2 x^{2} - 3 x + 2$ $P(x)=x^3-2x^2-3x+2$ is divided by

$(x - 1)$ $(x-1)$ , the Quotient Poly. is $Q (x)$ $Q(x)$ and the remainder $R .$ $R.$ Note

that, since the divisor $9 x - 1)$ $9x-1)$ is a Linear Poly., the Remainder has

to be a constant.

The well-known relation btwn. $P (x), Q (x), (x - 1) and, R$ $P(x), Q(x), (x-1) and, R$ is

given by, $P (x) = (x - 1) Q (x) + R,$ $P(x)=(x-1)Q(x)+R,$ i.e.,

$x^3-2x^2-3x+2=(x-1)Q(x)+R........................(star)$

Sub.ing, $x=1" in "(star), 1-2-3+2=(1-1)Q(x)+R :. R=-2.$

Then, sub.ing $R=-2" in "(star), x^3-2x^2-3x+2=(x-1)Q(x)-2,$

$or, x^3-2x^2-3x+4=(x-1)Q(x)$

$rArr Q(x)=(x^3-2x^2-3x+4)/(x-1)$

$=(x^3-x^2-x^2+x-4x+4)/(x-1)$

$={x^2(x-1)-x(x-1)-4(x-1)}/(x-1)$

$={cancel((x-1))(x^2-x-4)}/cancel((x-1))$

$:. Q(x)=x^2-x-4$ .

Enjoy Maths.!

How do you divide (x^3-2x^2-3x+2) / (x-1)x3−2x2−3x+2x−1(x^3-2x^2-3x+2) / (x-1)?