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Byju's Answer

Standard VIII

Mathematics

Factorisation by Common Factors

Consider the ...

Question

Consider the following statements :
1. $x - 2$ is a factor of $x^{3} - 3 x^{2} + 4 x - 4$
2. $x + 1$ is a factor of $2 x^{3} + 4 x + 6$
3. $x - 1$ is a factor of $x^{6} - x^{5} + x^{4} - x^{3} + x^{2} - x + 1$
Of these statements

1 and 2 are correct

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1, 2 and 3 are correct

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2 and 3 are correct

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1 and 3 are correct

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Solution

The correct option is A 1 and 2 are correct
1. Remainder $= 2^{3} - 3 \times 2^{2} + 4 \times 2 - 4$
$= 8 - 12 + 8 - 4 = 0$
Hence $x - 2$ is a factor.
2. Remainder $= 2 (- 1)^{3} + 4 (- 1) + 6$
$= - 2 - 4 + 6 = 0$
Hence $x + 1$ is a factor.
3. Ramainder $= 1^{6} - 1^{5} + 1^{5} - 1^{3} + 1^{2} - 1 + 1 = 1$
Hence $x - 1$ is not a factor.
$∴$ Statements 1 and 2 are correct.

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Similar questions

Q. Consider the following statements
(1)

x - 2

is a factor of

x^{3} - 3 x^{2} + 4 x - 4

(2)

x + 1

is a factor of

2 x^{3} + 4 x + 6

(3)

x - 1

is a factor of

x^{5} + x^{4} - x^{3} + x^{2} - x + 1

.
Of these statement.

Q. Match the following.

	Column - I		Column - II
(P)	$(x^{2} + 5) (x^{3} + 3) + 5$	(1)	${- x}^{3} - 3 x^{2} + 3 x + 2$
(Q)	$(\frac{- 10}{3} x y^{3}) \times (\frac{6}{5} x^{3} y)$	(2)	$- x^{3} + x^{2} + 3 x - 6$
(R)	$(x^{3} - x^{2} - x - 2) - (2 x^{3} + 2 x^{2} - 4 x - 4)$	(3)	$x^{5} + 5 x^{3} + 3 x^{2} + 20$
(S)	$(x^{3} - x^{2} - x - 2) + (2 x^{2} - 2 x^{3} + 4 x - 4)$	(4)	$- 4 x^{4} y^{4}$

You are given $c o s x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} . . . . . .;$

$s i n x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} . . . . . .$

$t a n x = x + \frac{x^{3}}{3} + \frac{2. x^{5}}{15} . . . . . .$

Find the value of $lim x \to 0 \frac{x c o s x + s i n x}{x^{2} + t a n x}$

___

You are given $c o s x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} . . . . . .;$

$s i n x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} . . . . . .$

$t a n x = x + \frac{x^{3}}{3} + \frac{2. x^{5}}{15} . . . . . .$

Find the value of $lim x \to 0 \frac{x c o s x + s i n x}{x^{2} + t a n x}$

You are given
$cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} . . . . . .$ ;
$sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} . . . . . .$ ;
$tan x = x + \frac{x^{3}}{3} + \frac{2 x^{5}}{15} . . . . . .$
Then the value of
$lim x \to 0 \frac{x cos x + sin x}{x^{2} + tan x}$ is

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Consider the following statements :1. x−2 is a factor of x3−3x2+4x−42. x+1 is a factor of 2x3+4x+63. x−1 is a factor of x6−x5+x4−x3+x2−x+1Of these statements

The correct option is A 1 and 2 are correct1. Remainder =23−3×22+4×2−4=8−12+8−4=0Hence x−2 is a factor.2. Remainder=2(−1)3+4(−1)+6=−2−4+6=0Hence x+1 is a factor.3. Ramainder =16−15+15−13+12−1+1=1Hence x−1 is not a factor.∴ Statements 1 and 2 are correct.

Consider the following statements :
1. $x - 2$ is a factor of $x^{3} - 3 x^{2} + 4 x - 4$
2. $x + 1$ is a factor of $2 x^{3} + 4 x + 6$
3. $x - 1$ is a factor of $x^{6} - x^{5} + x^{4} - x^{3} + x^{2} - x + 1$
Of these statements