Question

On dividing 3x3-2x2-5x+5 by a polynomial p(x) the quotient and the remainder are x2+px+q and 3 respectively. find p(x)

Answer 1

$\dfrac{3x^2-2x^2-5x+2}{x^2+px+q}$

Step-by-step explanation:

We know that,

Dividend = Divisor × Quotient + Remainder

Here,

Dividend = 3x³ - 2x² -5x + 5

Divisor = p(x)

Quotient = x² + px + q

Remainder = 3

By substituting the values,

We get,

$3x^3-2x^2-5x+5 = p(x)\times (x^2+px+q) + 3$

$3x^3-2x^2-5x+5 - 3 = p(x)\times (x^2+px+q)$

$3x^3-2x^2-5x+2 = p(x)\times (x^2+px+q)$

$\implies p(x) = \frac{3x^3-2x^2-5x+2}{x^2+px+q}$

#Learn more :

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Answer 2

$p(x)=\frac{3x^3-2x^2-5x+2}{(x^2+px+q)}$

Solution:

Dividend = $(3x^3-2x^2-5x+5)$

Quotient = $(x^2+px+q)$

Remainder = 3

Divisor = p(x)

To find p(x):

Dividend = Divisor × Quotient + Remainder

$3x^3-2x^2-5x+5=p(x)\times(x^2+px+q)+3$

Subtract 3 on both sides of the equation.

$3x^3-2x^2-5x+5-3=p(x)\times(x^2+px+q)+3-3$

$3x^3-2x^2-5x+2=p(x)\times(x^2+px+q)$

Divide by $(x^2+px+q)$ on both side of the equation.

$\frac{3x^3-2x^2-5x+2}{(x^2+px+q)} =p(x)\times\frac{(x^2+px+q)}{(x^2+px+q)}$

$\frac{3x^3-2x^2-5x+2}{(x^2+px+q)} =p(x)\times1$

$\frac{3x^3-2x^2-5x+2}{(x^2+px+q)} =p(x)$

Hence,

$p(x)=\frac{3x^3-2x^2-5x+2}{(x^2+px+q)}$