Q: \$3x^2-x-4\$.

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\times -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)\$.

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) \$

Q: Factorise \$6x^2+5x-6\$.

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.

Q: Factorise:\$ax+bx -ay-by\$

\$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]

Q: Factorize: \$6xy - 4y + 6 - 9x\$

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)\$.

Q: Factorise \$ 4p^2-9q^2\$

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

Q: Factorise.\$x^{ 2 } + xy + 8x + 8y\$

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

Q: Factorise:\$15xy-6x+5y -2\$

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

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

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)\$

Question 1

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

Accepted Answer

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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 $3\times -4=-12$.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 $25a^{2} - 4b^{2} + 28bc - 49c^{2}$.

Accepted Answer

$ 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 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:$ax+bx -ay-by$

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 6

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 7

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 8

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 9

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 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)$