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Long Division Method to Divide Two Polynomials

4x^3 + x^2 + ...

Question

$4 x^{3} + x^{2} + 5 x + 8$ = $(x + 1)$ $\times$ .

A

(4 x^{2} - 3 x + 8)

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B

(x^{2} + 8 x - 3)

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C

(x^{2} - 3 x + 8)

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D

(4 x^{2} + 3 x - 8)

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Solution

The correct option is A $(4 x^{2} - 3 x + 8)$
The quotient obtained when ( $4 x^{3} + x^{2} + 5 x + 8$ ) is divided by ( $x + 1$ ) is the other factor.

$\begin{matrix} 4 x^{2} - 3 x + 8 x + 1 4 x^{3} + x^{2} + 5 x + 8 4 x^{3} + 4 x^{2} - --------- - - 3 x^{2} + 5 x - 3 x^{2} - 3 x - ---------- - 8 x + 8 8 x + 8 - ----- - 0 \end{matrix}$

Hence, the other factor is $4 x^{2} - 3 x + 8$ .

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Similar questions

Q. Factorize the following polynomials.
(i) (x² – x)² – 8 (x² – x) + 12
(ii) (x – 5)² – (5x – 25) – 24
(iii) (x² – 6x)² – 8 (x² – 6x + 8) – 64
(iv) (x² – 2x + 3) (x² – 2x + 5) – 35
(v) (y + 2) (y – 3) (y + 8) (y + 3) + 56
(vi) (y² + 5y) (y² + 5y – 2) – 24
(vii) (x – 3) (x – 4)² (x – 5) – 6

Q. Which of the following are quadratic equations?

(i) x² + 6x − 4 = 0
(ii)

\sqrt{3} x^{2} - 2 x + \frac{1}{2} = 0

x^{2} + \frac{1}{x^{2}} = 5

x - \frac{3}{x} = x^{2}

2 x^{2} - \sqrt{3 x} + 9 = 0

x^{2} - 2 x - \sqrt{x} - 5 = 0

(vii) 3x² − 5x + 9 = x² − 7x + 3
(viii)

x + \frac{1}{x} = 1

(ix) x² − 3x = 0
(x)

{(x + \frac{1}{x})}^{2} = 3 (x + \frac{1}{x}) + 4

(xi) (2x + 1) (3x + 2) = 6(x − 1) (x − 2)
(xii)

x + \frac{1}{x} = x^{2}, x \neq 0

(xiii) 16x² − 3 = (2x + 5) (5x − 3)
(xiv) (x + 2)³ = x³ − 4
(xv) x(x + 1) + 8 = (x + 2) (x − 2)

Q. The quotient when

4 x^{3} + x^{2} + 5 x + 8

(x + 1)

Q. The factors of the polynomial equation P(x) =

4 x^{3} + x^{2} + 5 x + 8

The factors of the polynomial equation P(x) = $4 x^{3} + x^{2} + 5 x + 8$ is

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