Factorize - x2-1-4x2-1-7x-7-2x

The correct option is D ( x+1/2x ) ( x+1/2x 7 ) #160; #160; x^2+14x^2+1 7x 72x =x^2+14x^2+2(x)(12x) 7(x+12x) =(x+12x)^2 7(x+12x) ...

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Byju's Answer

Standard IX

Mathematics

Factorisation of Quadratic Polynomials - Factor Theorem

Factorize : ...

Question

Factorize :
$x^{2} + \frac{1}{4 x^{2}} + 1 - 7 x - \frac{7}{2 x}$

(x + \frac{1}{3 x}) (x + \frac{1}{x} - 9)

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(x + \frac{1}{5 x}) (x + \frac{1}{2 x} - 1)

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(x + \frac{1}{x}) (x + \frac{1}{x} - 12)

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(x + \frac{1}{2 x}) (x + \frac{1}{2 x} - 7)

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Solution

The correct option is D $(x + \frac{1}{2 x}) (x + \frac{1}{2 x} - 7)$
$x^{2} + \frac{1}{4 x^{2}} + 1 - 7 x - \frac{7}{2 x}$
$= x^{2} + \frac{1}{4 x^{2}} + 2 (x) (\frac{1}{2 x}) - 7 (x + \frac{1}{2 x})$
$= (x + \frac{1}{2 x})^{2} - 7 (x + \frac{1}{2 x})$
$= (x + \frac{1}{2 x}) ((x + \frac{1}{2 x}) - 7)$

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Similar questions

Factorize $2 x^{2} - 7 x + 5 = 0$

Check whether the following are quadratic equations :
(i) ${(x + 1)}^{2} = 2 (x - 3)$
(ii) $x^{2} - 2 x = (- 2) (3 - x)$
(iii) $(x - 2) (x + 1) = (x - 1) (x + 3)$
(iv) $(x - 3) (2 x + 1) = x (x + 5)$
(v) $(2 x - 1) (x - 3) = (x + 5) (x - 1)$
(vi) $x^{2} + 3 x + 1 = {(x - 2)}^{2}$
(vii) ${(x + 2)}^{3} = 2 x (x^{2} - 1)$
(viii) $x^{3} - 4 x^{2} - x + 1 = {(x - 2)}^{3}$

Solve :

$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$

Q. Solve for

x \frac{x - 1}{2 x + 1} + \frac{2 x + 1}{x - 1} = 2

where

x \neq \frac{- 1}{2}, 1

Q. Match the equations in List 1 with the solutions in List 2

	List I		List II
A.	${log}_{0.25} {(x^{2} + 2 x - 8)}^{2}$ $- {log}_{0.5} (10 + 3 x - x^{2}) = 1$	1.	${- 3}$
B.	${log}_{2} (x^{2} + 7)$ $= 5 + {log}_{2} x$ $- \frac{6}{{log}_{2} (x + \frac{7}{x})}$	2.	${\frac{1}{4}}$
C.	${log}_{(1 - 2 x)} (6 x^{2} - 5 x + 1)$ $- {log}_{(1 - 3 x)} (4 x^{2} - 4 x + 1) = 2$	3.	${\frac{1}{6} (\sqrt{313} - 1), \frac{1}{2} (\sqrt{73} - 7)}$
D.	${log}_{10} (1 {+ x}^{2} - 2 x) + 1 - {log}_{10} (1 + x^{2}) =$ $2 {log}_{10} (1 - x)$	4.	${1, 7}$

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Factorize : x2+14x2+1−7x−72x

The correct option is D (x+12x)(x+12x−7) x2+14x2+1−7x−72x=x2+14x2+2(x)(12x)−7(x+12x)=(x+12x)2−7(x+12x)=(x+12x)((x+12x)−7)

Factorize :
$x^{2} + \frac{1}{4 x^{2}} + 1 - 7 x - \frac{7}{2 x}$