Q: Factorize 25a^{2} - 4b^{2} + 28bc - 49c^{2} .

25{ a^{ 2 } }-4{ b^{ 2 } }-28bc-49{ c^{ 2 } } \\ = { \left( { 5a } \right) ^{ 2 } }-{ \left( { 2b } \right) ^{ 2 } }+28bc-{ \left( { 7c } \right) ^{ 2 } } \\ = { \left( { 5a } \right) ^{ 2 } }-\left\{ { { { \left( { 2b } \right) }^{ 2 } }+2\times 2b\times 7c-{ { \left( { 7c } \right) }^{ 2 } } } \right\} \\ \because \left[ { { { \left( { a+b } \right) }^{ 2 } }={ a^{ 2 } }+{ b^{ 2 } }+2ab } \right] \\ = { \left( { 5a } \right) ^{ 2 } }-{ \left( { 2b+7c } \right) ^{ 2 } } \\ \because \left[ { { a^{ 2 } }-{ b^{ 2 } }=\left( { a-b } \right) \left( { a+b } \right) } \right] \\ =\left\{ { \left( { 5a } \right) -\left( { 2b+7c } \right) } \right\} \left\{ { \left( { 5a } \right) +\left( { 2b+7c } \right) } \right\} \\ =\left( { 5a-2b-7c } \right) \left( { 5a+2b+7c } \right)

Question 1

Factorize  25a^{2} - 4b^{2} + 28bc - 49c^{2} .

Accepted Answer

25{ a^{ 2 } }-4{ b^{ 2 } }-28bc-49{ c^{ 2 } } \ = { \left( { 5a } ight) ^{ 2 } }-{ \left( { 2b } ight) ^{ 2 } }+28bc-{ \left( { 7c } ight) ^{ 2 } } \ = { \left( { 5a } ight) ^{ 2 } }-\left\{ { { { \left( { 2b } ight)  }^{ 2 } }+2	imes 2b	imes 7c-{ { \left( { 7c } ight)  }^{ 2 } } } ight\}  \         \because \left[ { { { \left( { a+b } ight)  }^{ 2 } }={ a^{ 2 } }+{ b^{ 2 } }+2ab } ight]  \ = { \left( { 5a } ight) ^{ 2 } }-{ \left( { 2b+7c } ight) ^{ 2 } } \         \because \left[ { { a^{ 2 } }-{ b^{ 2 } }=\left( { a-b } ight) \left( { a+b } ight)  } ight]  \  =\left\{ { \left( { 5a } ight) -\left( { 2b+7c } ight)  } ight\} \left\{ { \left( { 5a } ight) +\left( { 2b+7c } ight)  } ight\}  \  =\left( { 5a-2b-7c } ight) \left( { 5a+2b+7c } ight)

Question 2

3x^2-x-4 .

Accepted Answer

In the given polynomial  3x^2-x-4 ,The first term is  3x^2  and its coefficient is  3 .The middle term is  -x  and its coefficient is  -1 .The last term is a constant term  -4 .Multiply the coefficient of the first term by the constant  3	imes -4=-12 .We now find the factor of  -12  whose sum equals the coefficient of the middle term, which is  -1  and then factorize the polynomial  3x^2-x-4  as shown below:  3x^{ 2 }-x-4\ =3x^{ 2 }+3x-4x-4\ =3x(x+1)-4(x+1)\ =(3x-4)(x+1) Hence,  3x^{ 2 }-x-4=(3x-4)(x+1) .

Question 3

Factorize:  6xy - 4y + 6 - 9x

Accepted Answer

Given polynomial is  6xy-4y+6-9x It can be factorised as  6xy-4y+6-9x=2y(3x-2)-3(3x-2) =(3x-2)(2y-3) Thus factors are  (3x-2),(2y-3) .

Question 4

Factorise  6x^2+5x-6 .

Accepted Answer

Given expression is  6x^2+5x-6 6x^2+9x-4x-6 3x(2x+3)-2(2x+3) (2x+3)(3x-2) .so option B is correct.

Question 5

Factorise  x^3-2x^2-x+2

Accepted Answer

Toppr Ask Answer

Question 6

Factorise   4p^2-9q^2

Accepted Answer

4p^{2}-9q^{2}   \Rightarrow (2p)^{2}-(3q)^{2}    [a^{2}-b^{2} = (a+b)(a-b)]   \Rightarrow (2p+3q)(2p-3q)

Question 7

Factorise: ax+bx -ay-by

Accepted Answer

ax+bx -ay-by  = (ax+ bx) -(ay + by) [Grouping the terms] = (a+b)x-(a+b)y = (a+b)(x-y) [Taking (a + b) common]

Question 8

Factorise: 15xy-6x+5y -2

Accepted Answer

15xy-6x+5y -2 =15xy+5y-6x-2 =5y(3x+1)-2(3x+1) =(3x+1)(5y-2)

Question 9

Factorise. x^{ 2 } + xy + 8x + 8y

Accepted Answer

x^2+x{y}+8 x+8 y =x(x+y)+8(x+y) =(x+8)(x+y)

Question 10

Factorise the expressions and divide them as directed. (y^2 + 7y + 10) \div (y + 5)

Accepted Answer

We have, (y^{ 2 }+7y+10)\div (y+5) =\dfrac { (y^{ 2 }+5y+2y+10) }{ (y+5) }  =\dfrac { y(y+5)+2(y+5) }{ (y+5) }  =\dfrac { (y+5)(y+2) }{ (y+5) }  =(y+2)