[2 Mark Question Answer]

Question 512 Marks

Find trinomial $ ($quadratic expression $), $ given below $,$ find whether it is factorisable or not. Factorise$, $if possible.$2x^2+ 2x - 75$

Answer

Given expression $: 2x^2+ 2x - 75$
Comparing with $ax^2 + bx + c,$ we get $a = 2, b = 2$, and $c = - 75$
$\therefore b^2 - 4ac = (2)^2 - 4(2)(-75) = 4 + 600 = 604,$ which is not a perfect square.
$\therefore 2x^2 + 2x - 75$ is not factorisable.

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Question 522 Marks

Find trinomial $($quadratic expression$), $ given below $,$ find whether it is factorisable or not. Factorise $,$ if possible.$2x^2- 7x - 15$

Answer

Given expression $ :2x^2- 7x - 15$
Comparing with $ax^2 + bx + c$, we get $a = 2, b = -7$, and $c = -15$
$\therefore b^2 - 4ac = (-7)^2 - 4(2)(-15) = 49 + 120 = 169,$ which is a perfect square.
$\therefore 2x^2 - 7x - 15$ is factorisable.
Now $, 2x^2 - 7x - 15$
$= 2x2 - 10x + 3x - 15$
$= 2x( x - 5 ) + 3( x - 5 )$
$= ( 2x + 3 )( x - 5 )$

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Question 532 Marks

Find trinomial $($quadratic expression$),$ given below$,$ find whether it is factorisable or not. Factorise$,$ if possible.$x^2- 3x - 54$

Answer

Given expression $: x^2 - 3x - 54$
Comparing with $ax^2 + bx + c$, we get $a = 1, b = -3$, and $c = - 54$
$\therefore b^2 - 4ac = (-3)^2 - 4(1)(-54) = 9 + 216 = 225,$ which is a perfect square.
$\therefore x^2 - 3x - 54$ is factorisable.
Now, $x^2 - 3x - 54 = x^2 - 9x + 6x - 54$
$= x( x - 9 ) + 6( x - 9 )$
$= ( x - 9 )( x + 6 )$

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Question 542 Marks

Factorise$ : (3x - 2y)^2+ 3 (3x - 2y) - 10$

Answer

$(3x - 2y)^2+ 3 (3x - 2y) - 10$
Assume that $3x - 2y = a$
Therefore, $(3x - 2y)^2+ 3 (3x - 2y) - 10$
$= a2 + 3a - 10$
$= a2 + 5a - 2a -10$
$= a( a + 5 ) -2 ( a + 5 )$
$= ( a + 5 )( a - 2 )$
$= ( 3x - 2y + 5 )( 3x - 2y - 2)$

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Question 552 Marks

Factorise$ : x^2+ 3x + 2 + ax + 2a$

Answer

$x^2 + 3x + 2 + ax + 2a$
$= x^2 + 2x + x + 2 + ax + 2a$
$= x( x + 2 )+1( x + 2 ) +a( x + 2 )$
$= ( x + 2 )( x + a + 1 )$

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Question 562 Marks

Factorise$ : 3a^2- 1 - 2a$

Answer

$3a^2 - 1 - 2a$
$= 3a2 - 2a - 1$
$= 3a2 - 3a + a - 1$
$= 3a( a - 1 ) + 1( a - 1 )$
$= ( 3a + 1 )( a - 1 )$

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Question 572 Marks

Factorise by the grouping method $:ab(x^2+ 1) + x (a^2+ b^2)$

Answer

$ab (x^2 + 1) + x (a^2+ b^2)= abx^2 + ab + a^2x + b^2x$
$= abx^2 + b^2x + a^2x + ab$
$= bx(ax + b) + a(ax + b)$
$= (ax + b)(bx + a).$

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Question 582 Marks

Factorise by the grouping method $: 16 (a + b)^2- 4a - 4b$

Answer

$16 (a + b)^2 - 4a - 4b$
$= 16(a+b)^2- 4(a+b)$
$= (a + b) [16 (a + b) - 4]$
$= 4(a+b)[4(a+b)- 1]$
$= 4(a+b) (4a+4b- 1)$

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Question 592 Marks

Factorise by the grouping method : $a^3+ a - 3a^2- 3$

Answer

$a^3+ a - 3a^2- 3$
$=a (a^2+ 1) -3(a^2+ 1)$
$=(a^2+ 1)(a -3).$

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Question 602 Marks

Factories by taking out common factors $:2x(a - b) + 3y(5a - 5b) + 4z(2b - 2a)$

Answer

$2x(a - b) + 3y(5a - 5b) + 4z(2b - 2a)$
$= 2x(a - b) + 15y(a - b) - 8z(a - b)$
$= (a - b)[2x + 15y - 8z]$

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Question 612 Marks

Factories by taking out common factors $ :ab(a^2+ b^2- c^2) -bc(c^2- a^2- b^2) + ca(a^2+ b^2- c^2)$

Answer

$ab(a^2+ b^2- c^2) -bc(c^2- a^2- b^2) + ca(a^2+ b^2- c^2)$
$=ab(a^2+ b^2- c^2) +bc(a^2+ b^2- c^2) + ca(a^2+ b^2- c^2)$
$= (a^2+ b^2- c^2)[ab+bc+ ca]$

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Question 622 Marks

Factories by taking out common factors $:xy(3x^2- 2y^2) -yz(2y^2- 3x^2) +zx(15x^2- 10y^2)$

Answer

$xy(3x^2- 2y^2) -yz(2y^2- 3x^2) +zx(15x^2- 10y^2)$
$=xy(3x^2- 2y^2) +yz(3x^2- 2y^2) +zx(15x^2- 10y^2)$
$=xy(3x^2- 2y^2) +yz(3x^2- 2y^2) + 5zx(3x^2- 2y^2)$
$= (3x^2- 2y^2)[xy+yz+ 5zx]$

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Question 632 Marks

Factorise using the grouping method $:x (6x - 5y) - 4 (6x - 5y)^2$

Answer

$x (6x - 5y) - 4 (6x - 5y)^2$
$= (6x - 5y)[x - 4(6x - 5y)]$
$[$Taking $(6x - 5y)$ common from the three terms$]$
$= (6x - 5y)(x - 24x + 20y)$
$= (6x - 5y)(-23x + 20y)$
$= (6x - 5y)(20y - 23x)$

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Question 642 Marks

Factorise by taking out the common factors $:2 (2x - 5y) (3x + 4y) - 6 (2x - 5y) (x - y)$

Answer

$2 (2x - 5y) (3x + 4y) - 6 (2x - 5y) (x - y)$
Taking $ (2x - 5y) $ common from both terms
$= (2x - 5y)[2(3x + 4y) - 6(x - y)]$
$= (2x - 5y)(6x + 8y - 6x + 6y)$
$= (2x - 5y)(8y + 6y)$
$= (2x - 5y)(14y)$
$= (2x - 5y)14y$

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Question 652 Marks

Factorise using the grouping method $ :m (x - 3y)^2+ n (3y - x) + 5x - 15y$

Answer

$m (x - 3y)^2+ n (3y - x) + 5x - 15y$
$= m (x - 3y)^2 - n (x - 3y) + 5(x - 3y)$
$[$Taking $(x - 3y)$ common from all the three terms$]$
$= (x - 3y) [m(x - 3y) - n + 5]$
$= (x - 3y)(mx - 3my - n + 5)$

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Question 662 Marks

Factorise using the grouping method $:a^2+ 4b^2- 3a + 6b - 4ab$

Answer

$a^2+ 4b^2- 3a + 6b - 4ab$
$= a^2+ 4b^2- 4ab - 3a + 6b$
$= a^2+ (2b)^2 - 2 \times a \times (2b) - 3(a - 2b) [As (a - b)^2= a^2- 2ab + b^2]$
$= (a - 2b)^2- 3(a - 2b)$
$= (a - 2b)[(a - 2b)- 3]$
$= (a - 2b)(a - 2b - 3)$

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Question 672 Marks

Factoriseusing the grouping method $:x^2+ y^2+ x + y + 2xy$

Answer

$x^2+ y^2+ x + y + 2xy$
$= ( x^2+ y^2 + 2xy ) + ( x + y ) [As (x + y)^2= x^2+ 2xy + y^2]$
$= ( x + y )^2+ ( x + y )$
$= ( x + y )( x + y + 1 )$

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Question 682 Marks

Factorise by the grouping method $:a^2+\frac{1}{a^2}-2-3 a+\frac{3}{a}$

Answer

$ a^2+\frac{1}{a^2}-2-3 a+\frac{3}{a}$
$ =\left(a-\frac{1}{a}\right)^2-3\left(a-\frac{1}{a}\right)$
$ =\left(a-\frac{1}{a}\right)\left[\left(a-\frac{1}{a}\right)-3\right]$
$ =\left(a-\frac{1}{a}\right)\left[a-\frac{1}{a}-3\right]$

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Question 692 Marks

Factorise by the grouping method $ : y^2- (a + b) y +ab$

Answer

$y^2 - (a + b) y + ab$
$= y2 - ay - by + ab$
$= y( y - a ) - b( y - a )$
$= ( y - a )( y - b )$

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Question 702 Marks

Factorise by the grouping method $: (2a-b)^2-10a + 5b$

Answer

$( 2a - b)^2 - 10a + 5b$
$= ( 2a - b )^2 - 5( 2a - b )$
$= ( 2a - b )( 2a - b - 5 )$

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Question 712 Marks

Factorise by the grouping method $ : a^2x^2+ (ax^2+ 1) x + a$

Answer

$a^2x^2+ (ax^2 + 1) x + a$
$= a^2x^2 + a + (ax^2 + 1) x$
$= a( ax^2 + 1) + x( ax^2 + 1)$
$= ( a + x )( ax^2 + 1 )$

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Question 722 Marks

Factorise by the grouping method $: (ax + by)^2+ (bx- ay)^2$

Answer

$(ax + by)^2+ (bx- ay)^2$
$= a^2x^2 + b^2y^2 + 2axby + b^2x^2 + a^2y^2 - 2bxay$
$= a^2x^2 + b^2y^2 + b^2x^2 + a^2y^2$
$= x^2( a^2 + b^2 ) + y^2( a^2 + b^2 )$
$= ( x^2 + y^2 )( a^2 + b^2 )$

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Question 732 Marks

Factorise by the grouping method $: a^2+ b -ab- a$

Answer

$a^2 + b - ab - a$
$= a2 - a + b - ab$
$= a( a - 1) + b( 1 - a )$
$= a(a - 1) - b(a - 1)$
$= (a -1)(a - b)$

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MATHEMATICS [2 Mark Question Answer]