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Byju's Answer

Standard VII

Mathematics

Long Division Method

The HCF of ...

Question

The HCF of $x^{4} - 11 x^{2} + 10, x^{2} - 5 x + 4 a n d x^{3} - 3 x^{2} + 3 x - 1$ is

x + 1

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x - 4

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x + 2

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x - 1

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Solution

The correct option is D $x - 1$
$x^{4} - 11 x^{2} + 10 = x^{4} - 10 x^{2} - x^{2} + 10$
= $x^{2} (x^{2} - 10) - 1 (x^{2} - 10)$
= $(x^{2} - 10) (x^{2} - 1)$
= $(x^{2} - 10) (x + 1) (x - 1)$
$x^{2} - 5 x + 4 = (x - 4) (x - 1)$
$x^{3} - 3 x^{2} + 3 x + 1 = {(x - 1)}^{3}$
$∴$ Reqd HCH $= (x - 1)$

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Similar questions

Now in each of the pairs of polynomials given below, check whether the first is a factor of the second:

(i) x + 1, x³ − 1

(ii) x − 1, x³ + 1

(iii) x + 1, x³ + 1

(iv) x² − 1, x⁴ − 1

(v) x − 1, x⁴ − 1

(vi) x + 1, x⁴ − 1

(vii) x − 2, x² − 5x + 1

(viii) x + 2, x² + 5x + 6

(ix)

(x) 1.3x − 2.6, x² − 5x + 6

Q. A polynomial

p (x)

is divided by

g (x)

, the obtained quotient

q (x)

and the remainder

r (x)

are given in the table. Find

p (x)

in each case.

Sl.	$p (x)$	$g (x)$	$q (x)$	$r (x)$
i	?	$x - 2$	$x^{2} - x + 1$	$4$
ii	?	$x + 3$	$2 x^{2} + x + 5$	$3 x + 1$
iii	?	$2 x + 1$	$x^{3} + 3 x^{2} - x + 1$	$0$
iv	?	$x - 1$	$x^{3} - x^{2} - x - 1$	$2 x - 4$
v	?	$x^{2} + 2 x + 1$	$x^{4} - 2 x^{2} + 5 x - 7$	$4 x + 12$

Q. Determine which of the following polynomials has

(x + 1)

a factor:
(i)

x^{3} + x^{2} + x + 1

(ii)

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

(iii)

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

(iv)

x^{3} - x^{2} - (2 + \sqrt{2}) x + \sqrt{2}

Q. x + 1 is a factor of the polynomial
(a) x³ + x²– x + 1
(b) x³ + x² + x + 1
(c) x⁴ + x³ + x² + 1
(d) x⁴ + 3x³ + 3x² + x + 1

Q. Which of the following are quadratic equations?

(i) x² + 6x − 4 = 0
(ii)

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

(iii)

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

(iv)

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

(v)

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

(vi)

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)

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The HCF of x4−11x2+10,x2−5x+4andx3−3x2+3x−1 is

The correct option is D x−1x4−11x2+10=x4−10x2−x2+10 = x2(x2−10)−1(x2−10) = (x2−10)(x2−1) = (x2−10)(x+1)(x−1)x2−5x+4=(x−4)(x−1)x3−3x2+3x+1=(x−1)3∴ Reqd HCH =(x−1)

The HCF of $x^{4} - 11 x^{2} + 10, x^{2} - 5 x + 4 a n d x^{3} - 3 x^{2} + 3 x - 1$ is

The correct option is D $x - 1$
$x^{4} - 11 x^{2} + 10 = x^{4} - 10 x^{2} - x^{2} + 10$
= $x^{2} (x^{2} - 10) - 1 (x^{2} - 10)$
= $(x^{2} - 10) (x^{2} - 1)$
= $(x^{2} - 10) (x + 1) (x - 1)$
$x^{2} - 5 x + 4 = (x - 4) (x - 1)$
$x^{3} - 3 x^{2} + 3 x + 1 = {(x - 1)}^{3}$
$∴$ Reqd HCH $= (x - 1)$