Math, asked by rohitmeena25118, 10 months ago

find the remainder when p(x)= x cube - 6x square + 14x - 3 is divided by (x) = 1 - 2x and verify the result by long division

Answers

Answered by mysticd

$We \:have \: p(x) = x^{3}-6x^{2}+14x-3 \\divided \: by \: g(x) = 1- 2x = -2x + 1$

Quotient:(-x²)/2 -(11x/4)-(67/8)

-2x+1)x³-6x²+14x-3(

****** x³-(x²/2)

_______________

********** (-11x²/2)+14x

********** (-11x²/2) -(11x/4)

__________________

**************(67x/4)-3

**************(67x/4)-(67/8)

_____________________

*************** (43/8)

____________________

Dividend p(x) = x³-6x²+14x-3,

Divisor g(x) = 1-2x ,

Quotient q(x) = (-x²)/2 - (11x)/4 - (67/8),

Remainder r(x) = 43/8

Division Algorithm:

p(x) = g(x) × q(x) + r(x)

$(1-2x)\Big(\frac{-x^{2}}{2} - \frac{11x}{4} - \frac{67}{8}\Big) + \frac{43}{8}$

$= \frac{-x^{2}}{2} - \frac{11x}{4} - \frac{67}{8} \\+x^{3} + \frac{11x^{2}}{42} + \frac{67x}{4}+\frac{43}{8}$

$= x^{3} -6x^{2}+14x-3 \\= p(x)$

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Answered by kailashmeena123rm

ANSWER

we know

divisor = quotient × divisor + remainder

p(x) = quotient ×(1-2x) + remainder

$\\$

here

p(x) = $x^3 - 6x^2+14x-3$

g(x) = 1-2x

$\\$

$x^3 - 6x^2+14x-3 =( 1-2x)× quotient + remainder$

$\\$

put x = 1/2

so that ( 1-2x)× quotient becomes 0

$\\$

we get

remainder = $\frac{21}{8}$

$\\$

proof by long division method

divide $x^3 - 6x^2+14x-3$ by 1-2x

see attachment

$\\ \\ \\$

find the remainder when p(x)= x cube - 6x square + 14x - 3 is divided by (x) = 1 - 2x and verify the result by long division​

Answers

ANSWER

find the remainder when p(x)= x cube - 6x square + 14x - 3 is divided by (x) = 1 - 2x and verify the result by long division