Question

on dividing the polynomial 2x³-5x²+8x-5 by a polynomial g(x) the quotient and the remainder are (2x-3) and (3x-2) respectively find g(x).​

Answer 1

$\large\underline{\sf{Solution-}}$

Let we assume that

$\rm :\longmapsto\:f(x) = {2x}^{3} - {5x}^{2} + 8x - 5$

$\rm :\longmapsto\:q(x) = 2x - 3$

$\rm :\longmapsto\:r(x) = 3x - 2$

We know,

$\boxed{ \rm{ Dividend = Divisor × Quotient + Remainder}}$

$\rm :\longmapsto\:f(x) = q(x) \times g(x) + r(x)$

$\rm :\longmapsto\:f(x) - r(x) = q(x) \times g(x)$

$\rm :\longmapsto\:g(x) = \dfrac{f(x) - r(x)}{q(x)}$

$\rm :\longmapsto\:g(x) = \dfrac{ {2x}^{3} - {5x}^{2} + 8x - 5 - (3x - 2)}{2x - 3}$

$\rm :\longmapsto\:g(x) = \dfrac{ {2x}^{3} - {5x}^{2} + 8x - 5 - 3x + 2}{2x - 3}$

$\rm :\longmapsto\:g(x) = \dfrac{ {2x}^{3} - {5x}^{2} + 5x - 3}{2x - 3}$

Now, using Long Division, we get

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\: {x}^{2} - x + 1\:\:}}}\\ {\underline{\sf{2x - 3}}}& {\sf{\: {2x}^{3}  -  {5x}^{2} + 5x - 3 \:\:}} \\{\sf{}}& \underline{\sf{ \:  -2{x}^{3} \: + 3 {x}^{2}   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:\:}} \\ {{\sf{}}}& {\sf{\: \:  \:  \:  \:  \:  \:  \:  \:  \: \: -  2{x}^{2} + 5x - 3 \:  \:  \:  \:   \:  \:  \:  \:\:}} \\{\sf{}}& \underline{\sf{\:\: \:  \:  \:  \:\:  \: \: \: \: 2{x}^{2}  - 3x  \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \:  \:  \:  \:  \:  \:    2x - 3  \:\:}} \\{\sf{}}& \underline{\sf{\: \:  \:  \:  \:  \: - 2x  + 3\:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \: \:  \:  \:  \:  \:  \:  \:  \:  0\:\:}}  \end{array}\end{gathered}\end{gathered}\end{gathered}$

Hence,

$\bf\implies \:g(x) = {x}^{2} - x + 1$

ADDITIONAL INFORMATION :-

1. For cubic polynomial

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$

2. For quadratic polynomial

$\red{\rm :\longmapsto\: \alpha , \beta \: are \: zeroes \: of \: a {x}^{2} + b {x} + c, \: then}$

$\boxed{ \bf{ \: \alpha + \beta = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta = \dfrac{c}{a}}}$

Answer 2

Given :- on dividing the polynomial 2x³-5x²+8x-5 by a polynomial g(x) the quotient and the remainder are (2x-3) and (3x-2) respectively find g(x). ?

Solution :-

we know that,

• Dividend = Divisor × Quotient + Remainder.

given

→ Dividend = 2x³-5x²+8x-5

→ Divisor = g(x)

→ Quotient = (2x - 3)

→ Remainder = (3x - 2)

so,

→ 2x³-5x²+8x-5 = g(x) * (2x - 3) + (3x - 2)

→ 2x³-5x²+8x-5 - 3x + 2 = g(x) * (2x - 3)

→ g(x) = 2x³ - 5x² + 5x - 3 / 2x - 3

dividing now, we get

2x - 3 ) 2x³ - 5x² + 5x - 3 ( x² - x + 1

2x³ - 3x²

-2x² + 5x

-2x² + 3x

2x - 3

2x - 3

0

therefore,

→ Quotient = g(x) = x² - x + 1 (Ans.)

Learn more :-

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https://brainly.in/question/22246812

. Find all the zeroes of the polynomial x4

– 5x3 + 2x2+10x-8, if two of its zeroes are 4 and 1.

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