Question 12 Marks
Verify : $x^3+y^3=(x+y)\left(x^2-x y+y^2\right)$
AnswerWe know that
$(x+y)^3=x^3+y^3+3 x y(x+y)\left\{\text { Using Identity }(a+b)^3=a^3+b^3+3 a b(a+b)\right\}$
$\Rightarrow x^3+y^3=(x+y)^3-3 x y(x+y)$
$\Rightarrow x^3+y^3=(x+y)\left\{(x+y)^2-3 x y\right\}$
$\Rightarrow x^3+y^3=(x+y)\left(x^2+2 x y+y^2-3 x y\right)\left\{\text { Using Identity }(a+b)^2=a^2+2 a b+b^2\right\}$
$\Rightarrow x^3+y^3=(x+y)\left(x^2-x y+y^2\right)$
View full question & answer→Question 22 Marks
Verify : $x^3-y^3=(x-y)\left(x^2+x y+y^2\right)$
AnswerWe know that
$(x-y)^3=x^3-y^3-3 x y(x-y)\left\{\text { Using Identity }(a-b)^3=a^3-b^3-3 a b(a-b)\right\}$
$\Rightarrow x^3-y^3=(x-y)^3+3 x y(x-y)$
$\Rightarrow x^3-y^3=(x-y)\left\{(x-y)^2+3 x y\right\}$
$\Rightarrow x^3-y^3=(x-y)\left(x^2-2 x y+y^2+3 x y\right)$
$\Rightarrow x^3-y^3=(x-y)\left(x^2+x y+y^2\right)$
View full question & answer→Question 32 Marks
Factorise : $27 p^3-\frac{1}{216}-\frac{9}{2} p^2+\frac{1}{4} p$.
Answer$27 p^3-\frac{1}{216}-\frac{9}{2} p^2+\frac{1}{4} p$
$=(3 p)^3-\left(\frac{1}{6}\right)^3-3(3 p)\left(\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)$
$=\left(3 p-\frac{1}{6}\right)^3$
(Using Identity $(a-b)^3=a^3-b^3-3 a b(a-b)$ )
$=\left(3 p-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)$
View full question & answer→Question 42 Marks
Factorise : $8 a^3+b^3+12 a^2 b+6 a b^2$
Answer$8 a^3+b^3+12 a^2 b+6 a b^2$
$=(2 a)^3+(b)^3+3(2 a)(b)(2 a+b)$
$=(2 a+b)^3$
(Using Identity $(a+b)^3=a^3+b^3+3 a b(a+b)$ )
$=(2 a+b)(2 a+b)(2 a+b)$
View full question & answer→Question 52 Marks
Evaluate the using suitable identity: $(998)^3$
Answer$(998)^3$
$=(1000-2)^3=(1000)^3-(2)^2-3(1000)(2)(1000-2)$
$\left(\text { Using Identity }(a-b)^3=a^3-b^3-3 a b(a-b)\right)$
$=1000000000-8-6000(1000-2)$
$=1000000000-8-6000000+12000$
$=994011992$
View full question & answer→Question 62 Marks
Using suitable identities Evaluate : $(102)^3$
Answer$(102)^3$
$=(100+2)^3=(100)^3+(2)^3+3(100)(2)(100+2)$
$\left(\text { Using Identity }(a+b)^3=a^3+b^3+3 a b(a+b)\right)$
$=1000000+8+600(100+2)=1000000+8+60000+1200$
$=1061208$
View full question & answer→Question 72 Marks
Write the cube in expanded form : $\left(x-\frac{2}{3} y\right)^{3}$
Answer$\left(x-\frac{2}{3} y\right)^3=x^3-\left(\frac{2}{3} y\right)^3-3(x)\left(\frac{2}{3} y\right)\left(x-\frac{2}{3} y\right)$
(Using Identity $(a-b)^3=a^3-b^3-3 a b(a-b)$ )
$=x^3-\frac{8}{27} y^3-2 x y\left(x-\frac{2}{3} y\right)=x^3-\frac{8}{27} y^3-2 x^2 y+\frac{4}{3} x y^2$
$=x^3-2 x^2 y+\frac{4}{3} x y^2-\frac{8}{27} y^3$
View full question & answer→Question 82 Marks
Factorize:
$2{x^2} + {y^2} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz$
Answer$2{x^2} + {y^2} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz$
We need to factorize the expression $2{x^2} + {y^2} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz$
The expression $2{x^2} + {y^2} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz$ can also be written as
${\left( { - \sqrt 2 x} \right)^2} + {\left( y \right)^2} + {\left( {2\sqrt 2 z} \right)^2} + 2 \times \left( { - \sqrt 2 x} \right) \times y + 2 \times y \times \left( {2\sqrt 2 z} \right) + 2 \times \left( {2\sqrt 2 z} \right) \times \left( { - \sqrt 2 x} \right).$
We can observe that, we can apply the identity ${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca$ with respect to the expression
${\left( { - \sqrt 2 x} \right)^2} + {\left( y \right)^2} + {\left( {2\sqrt 2 z} \right)^2} + 2 \times \left( { - \sqrt 2 x} \right) \times y + 2 \times y \times \left( {2\sqrt 2 z} \right) + 2 \times \left( {2\sqrt 2 z} \right) \times \left( { - \sqrt 2 x} \right)$,to get
${\left( { - \sqrt 2 x + y + 2\sqrt 2 z} \right)^2}$
Therefore, we conclude that after factorizing the expression $2{x^2} + {y^2} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz $ we get ${\left( { - \sqrt 2 x + y + 2\sqrt 2 z} \right)^2}$.
View full question & answer→Question 92 Marks
Evaluate the product without multiplying directly: $104 \times 96$
Answer$104 \times 96$
$104 \times 96{\text{ can also be written as}}\left( {100 + 4} \right)\left( {100 - 4} \right).$ We can observe that, we can apply the identity $\left( {x + y} \right)\left( {x - y} \right) = {x^2} - {y^2}$ with respect to the expression $\left( {100 + 4} \right)\left( {100 - 4} \right)$,to get
$\left( {100 + 4} \right)\left( {100 - 4} \right) = {\left( {100} \right)^2} - {\left( 4 \right)^2}$ $= 10000 - 16$
$= 9984$
Therefore, we conclude that the value of the product $104 \times 96 $ is $9984$
View full question & answer→Question 102 Marks
Evaluate the product without multiplying directly: $95 \times 96$
Answer$95 \times 96$
$95 \times 96{\text{ can also be written as}}\left( {100 - 5} \right)\left( {100 - 4} \right)$ We can observe that we can apply the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$Here $a = -5$ and $b = -4$
$\left( {100 - 5} \right)\left( {100 - 4} \right) = {\left( {100} \right)^2} + \left[ {\left( { - 5} \right) + \left( { - 4} \right)} \right]\left( {100} \right) + \left( { - 5} \right) \times \left( { - 4} \right)$
$= 10000 - 900 + 20$
$= 9120$ Therefore, we conclude that the value of the product $95 \times 96$ is $9120$
View full question & answer→Question 112 Marks
What is the possible expression for the dimension of the cuboid whose volume is $12k{y^2} + 8ky - 20k$
Answer${\text{Volume : 12}}k{y^2} + 8ky - 20k$
The expression ${\text{12}}k{y^2} + 8ky - 20k $ can also be written as $ k\left( {{\text{12}}{y^2} + 8y - 20} \right).$
$k\left( {{\text{12}}{y^2} + 8y - 20} \right) = k\left( {{\text{12}}{y^2} - 12y + 20y - 20} \right)$ $\, = k\left[ {12y\left( {y - 1} \right) + 20\left( {y - 1} \right)} \right]$
$ = k\left( {12y + 20} \right)\left( {y - 1} \right)$
$= 4k \times \left( {3y + 5} \right) \times \left( {y - 1} \right).$
Therefore, we can conclude that a possible expression for the dimension of a cuboid of volume ${\text{12}}k{y^2} + 8ky - 20k \ is \ 4k\ ,\left( {3y + 5} \right){\text{ and }}\left( {y - 1} \right)$
View full question & answer→Question 122 Marks
Give possible expression for the length and breadth of the rectangle, in which the area is $35{y^2} + 13y - 12$
AnswerArea : $35 y^2+13 y-12$
The expression $35 y^2+13 y-12$ can also be written as, $35 y^2+28 y-15 y-12$
$=35 y^2+28 y-15 y-12=7 y(5 y+4)-3(5 y+4)$
$=(7 y-3)(5 y+4)$
Therefore, we can conclude that a possible expression for the length and breadth of a rectangle of area $35 y^2+13 y-12$ is Length $=7 y-3$ and Breadth $=5 y+4$
View full question & answer→Question 132 Marks
Give possible expression for the length and breadth of the rectangle, in which the area is $25{a^2} - 35a + 12$
AnswerArea: $25 a^2-35 a+12$
The expression $25 a^2-35 a+12$ can also written as $25 a^2-15 a-20 a+12$
$25 a^2-15 a-20 a+12=5 a(5 a-3)-4(5 a-3)$
$=(5 a-4)(5 a-3)$
Therefore, we can conclude that a possible expression for the length and breadth of a rectangle of area $25 a^2-35 a+12$ is Length $=(5 a-4)$ and Breadth $=(5 a-3)$
View full question & answer→Question 142 Marks
Use suitable identity to find the product: $(3-2 x)(3+2 x)$
Answer$(3-2 x)(3+2 x)=(3)^2-(2 x)^2$
$\left\{\right.$ Using Identity $\left.a^2-b^2=(a-b)(a+b)\right\}$
$=9-4 x^2$
View full question & answer→Question 152 Marks
Without actually calculating the cube, find the value of $(-12)^3+(7)^3+(5)^3$.
Answer$a^3+b^3+c^3=3 a b c$
$\text { If } a+b+c=0$
Given: $a=-12, b=7, c=5$
And $a+b+c=-12+7+5=0$
Then,
$(-12)^3+(7)^3+(5)^3=3 \times-12 \times 7 \times 5$
$=-1260$
View full question & answer→Question 162 Marks
Use suitable identity to find the product: $(x+8)(x-10)$
Answer$(x+8)(x-10)$
$=(x+8)\{x+(-10)\}$
(Using Identity $\left.(x+a)(x+b)=x^2+(a+b) x+a b\right)$
$=x^2+\{8+(-10)\} x+(8)(-10)$
$=x^2-2 x-80$
View full question & answer→Question 172 Marks
Factorise :
$27 y^3+125 z^3$
Answer$27 y^3+125 z^3$
$\left.=(3 y)^3+(5 z)^3=(3 y+5 z)(3 y)^2-(3 y)(5 z)+(5 z)^2\right\}$
$=(3 y+5 z)\left(9 y^2-15 y z+25 z^2\right)$
View full question & answer→Question 182 Marks
Find the remainder when $x^3-a x^2+6 x-a$ is divided by $x - a$.
Answer$\text { Let } p(x)=x^3-a x^2+6 x-a$
$x-a=0$
$\Rightarrow x=a$
$\therefore \text { Remainder }=(a)^3-a(a)^2+6(a)-a$
$=a^3-a^3+6 a-a$
$=5 a$
View full question & answer→Question 192 Marks
Find the remainder when $ {x^3} + 3{x^2} + 3x + 1$ is divided by $x - \frac{1}{2}$
Answer$x - \frac{1}{2}$
We need to find the zero of the polynomial $x - \frac{1}{2}$
$\begin{gathered} x - \frac{1}{2} = 0{\text{ }} \hfill \\ \Rightarrow {\text{ }}x = \frac{1}{2} \hfill \\ \end{gathered} $
While applying the remainder theorem, we need to put the zero of the polynomial $x - \frac{1}{2}$ in the polynomial ${x^3} + 3{x^2} + 3x + 1$, to get
$p\left( x \right) = {x^3} + 3{x^2} + 3x + 1$
$p\left( {\frac{1}{2}} \right) = {\left( {\frac{1}{2}} \right)^3} + 3{\left( {\frac{1}{2}} \right)^2} + 3\left( {\frac{1}{2}} \right) + 1$
$ = \frac{1}{8} + 3\left( {\frac{1}{4}} \right) + \frac{3}{2} + 1$ = ${1 \over 8} + {3 \over 4} + {3 \over 2} + 1$
$= \frac{{1 + 6 + 12 + 8}}{8}$
$\, = \frac{{27}}{8}$
Therefore, we conclude that on dividing the polynomial ${x^3} + 3{x^2} + 3x + 1 by \ x - \frac{1}{2}$ we will get the remainder as $\frac{{27}}{8}$
View full question & answer→Question 202 Marks
Verify $x = - \frac{1}{2}$ are zeroes of the polynomial $p\left( x \right) = 2x + 1$
Answer$p\left( x \right) = 2x + 1,\, x = - \frac{1}{2}$
We need to check whether $p\left( x \right) = 2x + 1,\, x = - \frac{1}{2}$ is equal to zero or not, i.e., $p\left( {{-1 \over 2}} \right)$ is equal to 0 or not.
$p\left( { - \frac{1}{2}} \right) = 2\left( { - \frac{1}{2}} \right) + 1\,\, = - 1 + 1\,\, = 0$ Since $p\left( {{-1 \over 2}} \right)$ = 0
Therefore, $x = - \frac{1}{2}$is a zero of the polynomial $p\left( x \right) = 2x + 1$
View full question & answer→Question 212 Marks
Verify $x = - \frac{m}{l}$ are zeroes of the polynomial $p\left( x \right) = lx + m$
Answer$p\left( x \right) = lx + m,\, x = - \frac{m}{l}$ We need to check whether $p\left( x \right) = lx + m{\text{ at }}x = - \frac{m}{l}$ is equal to zero or not, i.e., $p\left( {{{ - m} \over l}} \right)$ is equal to zero or not.
$p\left( { - \frac{m}{l}} \right) = l\left( { - \frac{m}{l}} \right) + m\,\, = -m + m\,\, = 0$
Therefore, $x = - \frac{m}{l}$ is a zero of the polynomial $p\left( x \right) = lx + m$
View full question & answer→Question 222 Marks
Verify $x = 0$ are zeroes of the polynomial $p\left( x \right) = {x^2}$
Answer$p\left( x \right) = {x^2},\,\,\,\,x = 0$
We need to check whether $p\left( x \right) = {x^2}{\text{ at }}x = 0$ is equal to zero or not, i.e., $p\left( 0 \right)$ is equal to zero or not.
$p\left( 0 \right) = {\left( 0 \right)^2}\, = 0$
Therefore, we can conclude that $x = 0$ is a zero of the polynomial $p\left( x \right) = {x^2}$
View full question & answer→Question 232 Marks
Verify $x = - 1,2$ are zeroes of the polynomial $p\left( x \right) = \left( {x + 1} \right)\left( {x - 2} \right)$
Answer$p\left( x \right) = \left( {x + 1} \right)\left( {x - 2} \right),\, x = - 1,2$ We need to check whether $p\left( x \right) = \left( {x + 1} \right)\left( {x - 2} \right){\text{ at }}x = - 1,2$ is equal to zero or not, i.e., $p\left( { - 1} \right)$ and $p\left( 2 \right)$ is equal to zero or not.
At $x = - 1$, $p\left( { - 1} \right) = \left( { - 1 + 1} \right)\left( { - 1 - 2} \right)\,\, = \left( 0 \right)\left( { - 3} \right)\,\, = 0$
At $x = 2$, $p\left( 2 \right) = \left( {2 + 1} \right)\left( {2 - 2} \right)\,\, = \left( 3 \right)\left( 0 \right)\,\, = 0$
Therefore, $x = - 1,2$ are the zeros of the polynomial $p\left( x \right) = \left( {x + 1} \right)\left( {x - 2} \right)$
View full question & answer→Question 242 Marks
Verify $x = - 1,1$ are zeroes of the polynomial $p\left( x \right) = {x^2} - 1$
Answer$p\left( x \right) = {x^2} - 1,\, x = - 1,1$
We need to check whether $p\left( x \right) = {x^2} - 1{\text{ at }}x = - 1,1$ is equal to zero or not, i.e., $p\left( { - 1} \right)$ and $p\left( { 1} \right)$ is equal to zero or not.
At $x = - 1$
$p\left( { - 1} \right) = {\left( { - 1} \right)^2} - 1\,\, = 1 - 1\,\, = 0$
At $x = 1$
$p\left( 1 \right) = {\left( 1 \right)^2} - 1\,\, = 1 - 1\,\, = 0$
Therefore ,$x = - 1,1$ are the zeros of the polynomial $p\left( x \right) = {x^2} - 1$
View full question & answer→Question 252 Marks
Verify $x = \frac{4}{5}$ are zeroes of the polynomial $p\left( x \right) = 5x - \pi$
Answer$p\left( x \right) = 5x - \pi ,\, x = \frac{4}{5}$
We need to check whether $p\left( x \right) = 5x - \pi {\text{ at }}x = \frac{4}{5}$ is equal to zero or not, i.e., $p\left( {{4 \over 5}} \right)$ is equal to zero or not.
$p\left( {\frac{4}{5}} \right) = 5\left( {\frac{4}{5}} \right) - \pi \,\, = 4 - \pi$
Therefore, $x = \frac{4}{5}$ is not a zero of the polynomial $ p\left( x \right) = 5x - \pi $
View full question & answer→Question 262 Marks
Verify $x = - \frac{1}{3}$ are zeroes of the polynomial $p\left( x \right) = 3x + 1$
Answer$p\left( x \right) = 3x + 1, x = - \frac{1}{3}$
We need to check whether $p\left( x \right) = 3x + 1{\text{ at }}x = - \frac{1}{3}$ is equal to zero or not.
$p\left( { - \frac{1}{3}} \right) = 3x + 1 = 3\left( { - \frac{1}{3}} \right) + 1\, = - 1 + 1\, = 0$
Therefore, we can conclude that $x = - \frac{1}{3}$ is a zero of the polynomial $p\left( x \right) = 3x + 1$
View full question & answer→Question 272 Marks
Find $p(0), p(1)$ and $p(2)$ for the polynomial: $p(x) = (x – 1)(x + 1)$
Answer$p(x) = (x – 1)(x + 1)$
$\therefore$ $p(0) = (0 – 1)(0 + 1) = (–1)(1) = –1$
$p(1) = (1 – 1)(1 + 1) = (0)(2) = 0,$
$p(2) = (2 – 1)(2 + 1) = (1)(3) = 3$
View full question & answer→Question 282 Marks
Find $p(0), p(1)$ and $p(2)$ of the polynomial: $p ( t ) = 2 + t + 2 t ^ { 2 } - t ^ { 3 }$
AnswerAccording to the question,$p ( t ) = 2 + t + 2 t ^ { 2 } - t ^ { 3 }$
$p ( 0 ) = 2 + ( 0 ) + 2 ( 0 ) ^ { 2 } - ( 0 ) ^ { 3 } = 2$
$p ( 1 ) = 2 + ( 1 ) + 2 ( 1 ) ^ { 2 } - ( 1 ) ^ { 3 } = 2 + 1 + 2 - 1 = 4$
$p ( 2 ) = 2 + ( 2 ) + 2 ( 2 ) ^ { 2 } - ( 2 ) ^ { 3 } = 4 + 8 - 8 = 4$
View full question & answer→Question 292 Marks
Verify whether $2$ and $0$ are zeroes of the polynomial $x^2-2 x$
AnswerLet $p(x)=x^2-2 x$
Then $p(2)=2^2-4=4-4=0$
and $p(0)=0-0=0$
Hence, $2$ and $0$ are both zeroes of the polynomial $x^2-2 x$.
View full question & answer→Question 302 Marks
Factorise : $8 x^3+y^3+27 z^3-18 x y z$
AnswerHere, we have
$8 x^3+y^3+27 z^3-18 x y z$
$=(2 x)^3+y^3+(3 z)^3-3(2 x)(y)(3 z)$
$=(2 x+y+3 z)\left[(2 x)^2+y^2+(3 z)^2-(2 x)(y)-(y)(3 z)-(2 x)(3 z)\right]$
$=(2 x+y+3 z)\left(4 x^2+y^2+9 z^2-2 x y-3 y z-6 x z\right)$
This is the required factorisation.
View full question & answer→Question 312 Marks
Factorise: $8 x^3+27 y^3+36 x^2 y+54 x y^2$
AnswerThe given expression can be written as
$(2 x)^3+(3 y)^3+3\left(4 x^2\right)(3 y)+3(2 x)\left(9 y^2\right)$
$=(2 x)^3+(3 y)^3+3(2 x)^2(3 y)+3(2 x)(3 y)^2$
$=(2 x+3 y)^3\left(\text { Using Identity }(x+y)^3=x^3+y^3+3 x y(x+y)\right)$
$=(2 x+3 y)(2 x+3 y)(2 x+3 y)$
This is the required factorisation.
View full question & answer→Question 322 Marks
Evaluate: $(104)^3$ using a suitable identity.
Answer$(104)^3=(100+4)^3$
Using identity $(x+y)^3=x^3+3 x y(x+y)+y^3$
We get,
$(100+4)^3=(100)^3+3 \times 100 \times 4(100+4)+4^3$
$=10,00000+1,200 \times 104+64$
$=10,00,000+1,24,800+64$
$=11,24,864$
View full question & answer→Question 332 Marks
Write $(5 p-3 q)^3$ in the expanded form.
AnswerComparing the given expression with $(x-y)^3$,
we find that $x=5 p, y=3 q$.
So, using Identity $(x-y)^3=x^3-3 x^2 y+3 x y^2-y^3$,
we have: $(5 p-3 q)^3=(5 p)^3-(3 q)^3-3(5 p)(3 q)(5 p-3 q)$
$=125 p^3-27 q^3-225 p^2 q+135 p q^2$
This is the required expansion.
View full question & answer→Question 342 Marks
Write $(3 a+4 b)^3$ in the expanded form.
AnswerComparing the given expression with $(x+y)^3$,
we find that $x=3$ a and $y=4 b$.
So, using Identity $(x+y)^3=x^3+y^3+3 x y(x+y)$,
we have: $(3 a+4 b)^3=(3 a)^3+(4 b)^3+3(3 a)(4 b)(3 a+4 b)$
$=27 a^3+64 b^3+108 a^2 b+144 a b^2$
This is the required expansion.
View full question & answer→Question 352 Marks
Find the value of $q(y)=3 y^3-4 y+\sqrt{11}$ at $y=2$.
AnswerWe have, $q(y)=3 y^3-4 y+\sqrt{11}$
On put $y=2$ in $q(y)$, we get
$q(2)=3(2)^3-4(2)+\sqrt{11}$
$=3 \times 8-8+\sqrt{11}$
$=24-8+\sqrt{11}$
$=16+\sqrt{11}$
View full question & answer→Question 362 Marks
Factorise : $4 x^2+y^2+z^2-4 x y-2 y z+4 x z$
AnswerWe have,
$4 x^2+y^2+z^2-4 x y-2 y z+4 x z=(2 x)^2+(-y)^2+(z)^2+2(2 x)(-y)+2(-y)(z)+2(2 x)(z)$
$=[2 x+(-y)+z]^2$
$=(2 x-y+z)^2$
$=(2 x-y+z)(2 x-y+z)$
This is the required factorisation.
View full question & answer→Question 372 Marks
Factorise : $49 a^2+70 a b+25 b^2$
AnswerHere you can see that $49 a^2=(7 a)^2, 25 b^2=(5 b)^2, 70 a b=2(7 a)(5 b)$
Comparing with $x^2+2 x y+y^2$, we observe that $x=7 a$ and $y=5 b$.
Using Identity $(x+y)^2=x^2+2 x y+y$, we get
$49 a^2+70 a b+25 b^2=(7 a+5 b)^2=(7 a+5 b)(7 a+5 b)$
This is the required factorisation.
View full question & answer→Question 382 Marks
Evaluate $105 \times 106$ without multiplying directly.
AnswerFirstly, write 105 as $100+5$ and 106 as $100+6$.
$\therefore 105 \times 106=(100+5) \times(100+6)$
$=(100)^2+(5+6) 100+(5 \times 6)$
${\left[\because(x+a)(x+b)=x^2+(a+b) x+a b\right]}$
$=10000+11 \times 100+30$
$=10000+1100+30=11130$
View full question & answer→Question 392 Marks
Factorise $y^2-5 y+6$ by using the Factor Theorem.
AnswerLet $p(y)=y^2-5 y+6$.
The factors of 6 are 1,2 and 3 .
Now, $p(2)=22-(5 \times 2)+6=0$
So, $y-2$ is a factor of $p(y)$.
Also, $p(3)=32-(5 \times 3)+6=0$
So, $y-3$ is also a factor of $y^2-5 y+6$
Therefore, $y^2-5 y+6=(y-2)(y-3)$
This is the required factorisation.
View full question & answer→Question 402 Marks
Factorise $6 x^2+17 x+5$ by splitting the middle term
AnswerIf we can find two numbers $p$ and $q$ such that $p+q=17$ and $p q=6 \times 5=30$, then we can get the factors So, let us look for the pairs of factors of $30 $. Some are $1$ and $30,2$ and $15,3$ and $10,5$ and $6 $. Of these pairs, $2$ and $15$ will give us $\mathrm{p}+\mathrm{q}=17$.
$\text { So, } 6 x^2+17 x+5$
$=6 x^2+(2+15) x+5$
$=6 x^2+2 x+15 x+5$
$=2 x(3 x+1)+5(3 x+1)$
$=(3 x+1)(2 x+5)$
This is the required factorisation.
View full question & answer→Question 412 Marks
Find the value of $k$, if $x-1$ is a factor of $4 x^3+3 x^2-4 x+k$.
AnswerAs $x-1$ is a factor of $p(x)=4 x^3+3 x^2-4 x+k$, therefore
$p(1)=0$
Now, $p(1)=4(1)^3+3(1)^2-4(1)+k$
So, $4+3-4+k=0$
i.e., $k=-3$
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