Question 11 Mark
Factories : $64 a^3-27 b^3-144 a^2 b+108 a b^2$
Answer$64 a^3-27 b^3-144 a^2 b+108 a b^2$
$=(4 a)^3-(3 b)^3-3(4 a)(3 b)(4 a-3 b)$
$=(4 a-3 b)^3$
(Using Identity $(a-b)^3=a^3-b^3-3 a b(a-b)$ )
$=(4 a-3 b)(4 a-3 b)(4 a-3 b)$
View full question & answer→Question 21 Mark
Factorise : $27-125 a^3-135 a+225 a^2$
Answer$27-125 a^3-135 a+225 a^2$
$=(3)^3-(5 a)^3-3(3)(5 a)(3-5 a)$
$=(3-5 a)^3$
(Using Identity $(a-b)^3=a^3-b^3-3 a b(a-b)$ )
$=(3-5 a)(3-5 a)(3-5 a)$
View full question & answer→Question 31 Mark
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 41 Mark
Evaluate using suitable identity: ${\left( {99} \right)^3}$
Answer${\left( {99} \right)^3}$
${\left( {99} \right)^3}{\text{ can also be written as}}{\left( {100 - 1} \right)^3}.$
Using identity, ${\left( {x - y} \right)^3} = {x^3} - {y^3} - 3xy\left( {x - y} \right)$
${\left( {100 - 1} \right)^3} = {\left( {100} \right)^3} - {\left( 1 \right)^3} - 3 \times 100 \times 1\left( {100 - 1} \right)$
$= 1000000-1-300(99)$
$= 999999-29700$
$= 970299.$
View full question & answer→Question 51 Mark
Write ${\left( {\frac{3}{2}x + 1} \right)^3}$ in expanded form.
Answer${\left( {\frac{3}{2}x + 1} \right)^3}$
We know that ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$
${\left( {\frac{3}{2}x + 1} \right)^3} = {\left( {\frac{3}{2}x} \right)^3} + {\left( 1 \right)^3} + 3 \times \frac{3}{2}x \times 1\left( {\frac{3}{2}x + 1} \right)$
$= \frac{{27}}{8}{x^3} + 1 + \frac{9}{2}x\left( {\frac{3}{2}x + 1} \right)$$ = \frac{{27}}{8}{x^3} + \frac{{27}}{4}{x^2} + \frac{9}{2}x + 1.$
Therefore, the expansion of the expression ${\left( {\frac{3}{2}x + 1} \right)^3}$ is $\frac{{27}}{8}{x^3} + \frac{{27}}{4}{x^2} + \frac{9}{2}x + 1$
View full question & answer→Question 61 Mark
Write $(2 a-3 b)^3$ in the expanded form
Answer$(2 a-3 b)^3$
$=(2 a)^3-(3 b)^3-3(2 a)(3 b)(2 a-3 b)$
(Using Identity $(a-b)^3=a^3-b^3-3 a b(a-b)$ )
$=8 a^3-27 b^3-18 a b(2 a-3 b)$
$=8 a^3-27 b^3-36 a^2 b+54 a b^2$
View full question & answer→Question 71 Mark
Using suitable identity Expand: $\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}$
Answer$\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}$ = $\left[\frac{1}{4} a+\left(-\frac{1}{2} b\right)+1\right]^{2}$
(Using Identity $\left.(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a\right)$
= $\left(\frac{1}{4} a\right)^{2}+\left(-\frac{1}{2} b\right)^{2}$ + $+(1)^{2}+2\left(\frac{1}{4} a\right)\left(-\frac{1}{2} b\right)+2\left(-\frac{1}{2} b\right)(1)+2(1)\left(\frac{1}{4} a\right)$
= $\frac{1}{16} a^{2}+\frac{1}{4} b^{2}+1-\frac{1}{4} a b-b+\frac{1}{2} a$
View full question & answer→Question 81 Mark
Factorise :$4 x^2+9 y^2+16 z^2+12 x y-24 y z-16 x z$
Answer$4 x^2+9 y^2+16 z^2+12 x y-24 y z-16 x z$
$=(2 x)^2+(3 y)^2+(-4 z)^2+2(2 x)(3 y)$
$+2(3 y)(-4 z)+2(-4 z)(2 x)$
$=\{2 x+3 y+(-4 z)\}^2=(2 x+3 y-4 z)^2$
$=(2 x+3 y-4 z)(2 x+3 y-4 z)$
View full question & answer→Question 91 Mark
Expand using appropriate identity: $(-2 x+5 y-3 z)^2$
Answer$(-2 x+5 y-3 z)^2$
$=\{(-2 x)+5 y+(-3 z)\}^2$
$=(-2 x)^2+(5 y)^2+(-3 z)^2+2(-2 x)(5 y)+2(5 y)(-3 z)+2(-3 z)(-2 x)$
$=4 x^2+25 y^2+9 z^2-20 x y-30 y z+12 z x$
View full question & answer→Question 101 Mark
Expand using appropriate identity: $(3 a-7 b-c)^2$
Answer
$(3 a-7 b-c)^2=\{3 a+(-7 b)+(-c)\}^2$
$=(3 a)^2+(-7 b)^2+(-c)^2+2(3 a)(-7 b)+2(-7 b)(-c)+2(-c)(3 a)$
$=9 a^2+49 b^2+c^2-42 a b+14 b c-6 c a$
View full question & answer→Question 111 Mark
Expand using suitable identity: $(-2 x+3 y+2 z)^2$
Answer$(-2 x+3 y+2 z)^2$
$=\{(-2 x)+3 y+(2 z)\}^2$
$=(-2 x)^2+(3 y)^2+(2 z)^2+2(-2 x)(3 y)+2(3 y)(2 z)+2(2 z)(-2 x)$
$\left(\text { Using Identity }(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a\right)$
$=4 x^2+9 y^2+4 z^2-12 x y+12 y z-8 z x$
View full question & answer→Question 121 Mark
Expand using suitable identity: $(2 x-y+z)^2$
Answer$(2 x-y+z)^2$
$=\{2 x+(-y)+z\}^2$
$=(2 x)^2+(-y)^2+(z)^2+2(2 x)(-y)+2(-y)(z)+2(z)(2 x)$
$\left(\text { Using Identity }(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a\right)$
$=4 x^2+y^2+z^2-4 x y-2 y z+4 z x$
View full question & answer→Question 131 Mark
Expand using suitable identity: $(x+2 y+4 z)^2$
Answer$(x+2 y+4 z)^2$
$=(x)^2+(2 y)^2+(4 z)^2+2(x)(2 y)+2(2 y)(4 z)+2(4 z)(x)$
(Using Identity $(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a$ )
$=x^2+4 y^2+16 z^2+4 x y+16 y z+8 z x$
View full question & answer→Question 141 Mark
Factorize the using appropriate identity: ${x^2} - \frac{{{y^2}}}{{100}}$
Answer${x^2} - \frac{{{y^2}}}{{100}}$
We can observe that we can apply the identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$
$ \Rightarrow {\left( x \right)^2} - {\left( {\frac{y}{{10}}} \right)^2} = \left( {x + \frac{y}{{10}}} \right)\left( {x - \frac{y}{{10}}} \right).$
View full question & answer→Question 151 Mark
Factorize the using appropriate identity: $4{y^2} - 4y + 1$
Answer$4{y^2} - 4y + 1 = {\left( {2y} \right)^2} - 2 \times 2y \times 1 + {\left( 1 \right)^2}$ We can observe that we can apply the identity ${\left( {x - y} \right)^2} = {x^2} - 2xy + {y^2} \Rightarrow {\left( {2y} \right)^2} - 2 \times 2y \times 1 + {\left( 1 \right)^2} = {\left( {2y - 1} \right)^2}$.
View full question & answer→Question 161 Mark
Factorize the using appropriate identity: $9{x^2} + 6xy + {y^2}$
Answer $9{x^2} + 6xy + {y^2}{\text{ = }}{\left( {3x} \right)^2} + 2 \times 3x \times y + {\left( y \right)^2}$
We can observe that we can apply the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$
$ \Rightarrow {\left( {3x} \right)^2} + 2 \times 3x \times y + {\left( y \right)^2} = {\left( {3x + y} \right)^2}.$
View full question & answer→Question 171 Mark
Evaluate the product without multiplying directly: $103 \times 107$
Answer$103 \times 107$
$103 \times 107$ can also be written as $(100+3)(100+7)$.
We can observe that we can apply the identity $(x+a)(x+b)=x^2+(a+b) x+a b$
$(100+3)(100+7)=(100)^2+(3+7)(100)+3 \times 7$
$=10000+1000+21$
$=11021$
Therefore, we conclude that the value of the product $103 \times 107$ is $11021 .$
View full question & answer→Question 181 Mark
What are the possible expression for the dimensions of the cuboid whose volume are: $3{x^2} - 12x$
Answer${\text{Volume : }}3{x^2} - 12x$
${\text{The expression }}3{x^2} - 12x{\text{ can also be written as}}\,3 \times x \times \left( {x - 4} \right).$
Therefore, we can conclude that a possible expression for the dimensions of a cuboid of volume $3{x^2} - 12x$ is $3,x{\text{ and }}\left( {x - 4} \right)$
View full question & answer→Question 191 Mark
Without actually calculating the cube, find the value of $(28)^3+(-15)^3+(-13)^3$
Answer$(28)^3+(-15)^3+(-13)^3$
$=3(28)(-15)(-13)(\because(28)+(-15)+(-13)=0)$
$=16380$
View full question & answer→Question 201 Mark
Use suitable identity to find the product $\left( {{y^2} + \frac{3}{2}} \right)\left( {{y^2} - \frac{3}{2}} \right)$
Answer$\left( {{y^2} + \frac{3}{2}} \right)\left( {{y^2} - \frac{3}{2}} \right)$
We know that $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$
We need to apply the above identity to find the product $\left( {{y^2} + \frac{3}{2}} \right)\left( {{y^2} - \frac{3}{2}} \right)$
Here $a = {y^2}$ and $b = {3 \over 2}$
$\left( {{y^2} + \frac{3}{2}} \right)\left( {{y^2} - \frac{3}{2}} \right) = {\left( {{y^2}} \right)^2} - {\left( {\frac{3}{2}} \right)^2}$
$= {y^4} - \frac{9}{4}.$
Therefore, we conclude that the product $\left( {{y^2} + \frac{3}{2}} \right)\left( {{y^2} - \frac{3}{2}} \right)$is $\left( {{y^4} - \frac{9}{4}} \right)$
View full question & answer→Question 211 Mark
If $x+y+z=0$ then show that $x^3+y^3+z^3=3 x y z$.
AnswerWe know that
$x^3+y^3+z^3-3 x y z=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$
(Using Identity $\left.a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)\right)$
$=(0)\left(x^2+y^2+z^2-x y-y z-z x\right)(\because x+y+z=0)$
$=0$
$\Rightarrow x^3+y^3+z^3=3 x y z$
View full question & answer→Question 221 Mark
Use suitable identity to find the product: $(3 x+4)(3 x-5)$
Answer$(3 x+4)(3 x-5)$
$=(3 x+4)\{3 x+(-5)\}$
$\left\{\right.$ Using identity $\left.(x+a)(x+b)=x^2+(a+b) x+a b\right\}$
$=(3 x)^2+\{4+(-5)\}(3 x)+(4)(-5)$
$=9 x^2-3 x-20$
View full question & answer→Question 231 Mark
Factorise : $27 x^3+y^3+z^3-9 x y z$.
Answer$27 x^3+y^3+z^3-9 x y z$
$(3 x)^3+(y)^3+(z)^3-3(3 x)(y)(z)$
$=(3 x+y+z)\left\{(3 x)^2+(y)^2+(z)^2-(3 x)(y)-(y)(z)-(z)(3 x)\right\}$
(Using Identity $\left.a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)\right)$
$=(3 x+y+z)\left(9 x^2+y^2+z^2-3 x y-y z-3 z x\right)$
View full question & answer→Question 241 Mark
Factorise :$64 m^3-343 n^3$
Answer$64 m^3-343 n^3$
$=(4 m)^3-(7 n)^3=(4 m-7 n)\left\{(4 m)^2+(4 m)(7 n)+(7 n)^2\right\}$
$=(4 m-7 n)\left(16 m^2+28 m n+49 n^2\right)$
View full question & answer→Question 251 Mark
Check whether $7 + 3x $ is a factor of $ 3{x^3} + 7x$.
AnswerWe know that if the polynomial $7+3x$ is a factor of $ 3{x^3} + 7x$, then on dividing the polynomial $3{x^3} + 7x$ by $7+3x$, we must get the remainder as $0.$
We need to find the zero of the polynomial $7 + 3x$
$\begin{gathered} 7 + 3x = 0 \hfill \\ \Rightarrow {\text{ }}x = - \frac{7}{3} \hfill \\ \end{gathered} $
While applying the remainder theorem, we need to put the zero of the polynomial $7+3x$ in the polynomial $ 3{x^3} + 7x$,to get
$\,\,p\left( x \right) = 3{x^3} + 7x$
$p\left( {{{ - 7} \over 3}} \right)$ $= 3{\left( { - \frac{7}{3}} \right)^3} + 7\left( { - \frac{7}{3}} \right)\,\, = 3\left( { - \frac{{343}}{{27}}} \right) - \frac{{49}}{3}$
$ = - \frac{{343}}{9} - \frac{{49}}{3}\,\, = \frac{{ - 343 - 147}}{9}$
$= \frac{{ - 490}}{9}.$
We conclude that on dividing the polynomial $3{x^3} + 7x $ by $7 + 3x$, we will get the remainder as $\frac{{ - 490}}{9}$ which is not $0.$
Therefore, we conclude that $7 + 3x$ is not a factor of $ 3{x^3} + 7x$
View full question & answer→Question 261 Mark
Find the remainder when $x^3+3 x^2+3 x+1$ is divided by $5+2 x$.
Answer$\text { Let } p(x)=x^3+3 x^2+3 x+1$
$5+2 x=0$
$\Rightarrow 2 x=-5$
$\Rightarrow x=-\frac{5}{2}$
$=\left(-\frac{5}{2}\right)^3+3\left(-\frac{5}{2}\right)^2+3\left(-\frac{5}{2}\right)+1$
$=-\frac{125}{8}+\frac{75}{4}-\frac{15}{2}+1=-\frac{27}{8}$
View full question & answer→Question 271 Mark
Find the remainder when $x^3+3 x^2+3 x+1$ is divided by $x+\pi$
AnswerLet $\mathrm{p}(\mathrm{x})=\mathrm{x}^3+3 \mathrm{x}^2+3 \mathrm{x}+1$
$x+\pi=0$
$\Rightarrow \mathrm{x}=-\pi$
$\therefore$ Remainder $=(-\pi)^3+3(-\pi)^2+3(-\pi)+1=-\pi^3+3 \pi^2-3 \pi+1$
View full question & answer→Question 281 Mark
Find the remainder when ${x^3} + 3{x^2} + 3x + 1$ is divided by $x + 1$
Answer$x + 1$
We need to find the zero of the polynomial $x + 1$
$x + 1 = 0{\text{ }} \Rightarrow {\text{ }}x = - 1$
While applying the remainder theorem, we need to put the zero of the polynomial $x + 1$ in the polynomial
${x^3} + 3{x^2} + 3x + 1$, to get
$p\left( x \right) = {x^3} + 3{x^2} + 3x + 1$
$p\left( -1 \right) = {\left( { - 1} \right)^3} + 3{\left( { - 1} \right)^2} + 3\left( { - 1} \right) + 1=-1+3-3+1
= 0$
Therefore, we conclude that on dividing the polynomial ${x^3} + 3{x^2} + 3x + 1$ by $x + 1,$ we will get the remainder as $0.$
View full question & answer→Question 291 Mark
Find the zero of the polynomial in $p(x) = 3x$
Answer$p(x) = 3x$
put, $p(x) = 0$
$3x = 0$
$x = 0$
Therefore, $x = 0$ is a zero of the polynomial $p(x) = 3x.$
View full question & answer→Question 301 Mark
Find the zero of the polynomial in $p(x) = 3x - 2.$
AnswerWe have,
$p(x) = 3x - 2$
$p(x) = 0$
$3x – 2 = 0$
$x = \frac{2}{3}$ Therefore, $x = \frac{2}{3}$ is a zero of the polynomial $p(x) = 3x - 2$
View full question & answer→Question 311 Mark
Find the zero of the polynomial in $p(x) = 2x + 5$
AnswerThe given polynomial is,
$P(x) = 2x + 5$
put $P(x) = 0$
$2x + 5 = 0$
$2x = -5$
$x =\frac{-5}{2}$
Therefore, $x =\frac{-5}{2}$ is a zero of the polynomial $p(x) = 2x + 5$
View full question & answer→Question 321 Mark
Find the zero of the polynomial in $p(x) = x – 5$
Answer$p(x) = x - 5$
put $p(x)=0$
$x - 5 = 0$
$x = 5$
Therefore, $x = 5$ is a zero of the polynomial $p(x) = x - 5$
View full question & answer→Question 331 Mark
Find the zero of the polynomial in $p(x) = x + 5.$
AnswerZero of a polynomial means the value of the variable at which the polynomial becomes zero.
$p(x) = x + 5p(x)=0$
$x + 5 = 0$
$x = -5$
Thus, $x = -5$ is a zero of the polynomial $p(x) = x + 5.$
View full question & answer→Question 341 Mark
Find $p(0), p(1)$ and $p(2)$ for the polynomial: $p(x)=x^3$
Answer$p(x)=x^3$
$\therefore p(0)=(0)^3=0,$
$p(1)=(1)^3=1$
$p(2)=(2)^3=8$
View full question & answer→Question 351 Mark
Find $p(0), p(1)$ and $p(2)$ for the polynomial: $p(y)=y^2-y+1$
Answer$p(y)=y^2-y+1$
$\therefore p(0)=(0)^2-(0)+1=1$
$p(1)=(1)^2-(1)+1=1$
$p(2)=(2)^2-(2)+1=4-2+1=3$
View full question & answer→Question 361 Mark
Find the value of the polynomial $5 x-4 x^2+3$ at $x=2$
AnswerLet $f(x)=5 x-4 x^2+3$
The value of $f(x)$ at $x=2$
$f(2)=5(2)-4(2)^2+3$
$=10-16+3=-3$
View full question & answer→Question 371 Mark
Find the value of the polynomial $5 x-4 x^2+3$ at $x=-1$
AnswerLet $f(x)=5 x-4 x^2+3$
The value of $f(x)$ at $x=-1$
$f(-1)=5(-1)-4(-1)^2+3$
$=-5-4+3=-6$
View full question & answer→Question 381 Mark
Find the value of the polynomial $5 x-4 x^2+3$ at $x=0$
AnswerLet $f(x)=5 x-4 x^2+3$
We need to substitute $x=0$ in the polynomial $f(x)=5 x-4 x^2+3$ to get the value of given polynomial.
$f(0)=5(0)-4(0)^2+3$
$=0-0+3$
$=3$
Therefore, we conclude that at $x=0$, the value of the polynomial $5 x-4 x^2+3$ is $3 .$
View full question & answer→Question 391 Mark
Classify as linear, quadratic and cubic polynomials:
$7{x^3}$
Answer$7{x^3}$
We can observe that the degree of the polynomial $7{x^3}$ is $3.$
Therefore, we can conclude that the polynomial $7{x^3}$ is a cubic polynomial.
View full question & answer→Question 401 Mark
Classify as linear, quadratic and cubic polynomials:
${r^2}$
Answer${r^2}$
We can observe that the degree of the polynomial ${r^2}$ is $2.$
Therefore, we can conclude that the polynomial ${r^2}$ is a quadratic polynomial.
View full question & answer→Question 411 Mark
Classify as linear, quadratic and cubic polynomials:
$3t$
Answer$3t$
We can observe that the degree of the polynomial $\left( {3t} \right)$ is $1.$ Therefore, we can conclude that the polynomial $3t$ is a linear polynomial.
View full question & answer→Question 421 Mark
Classify as linear, quadratic and cubic polynomials:
$1 + x{\text{ }}$
Answer$1 + x{\text{ }}$
We can observe that the degree of the polynomial $\left( {1 + x{\text{ }}} \right)$ is $1.$
Therefore, we can conclude that the polynomial $1 + x{\text{ }}$ is a linear polynomial.
View full question & answer→Question 431 Mark
Classify as linear, quadratic and cubic polynomials:
$y + {y^2} + 4$
Answer$y + {y^2} + 4$
We can observe that the degree of the polynomial $y + {y^2} + 4$ is $2.$
Therefore, the polynomial $y + {y^2} + 4$ is a quadratic polynomial.
View full question & answer→Question 441 Mark
Classify the linear, quadratic and cubic polynomials: $x - {x^3}$
Answer$x - {x^3}$
We can observe that the degree of the polynomial $x - {x^3}$ is $3.$
Therefore, we can conclude that the polynomial $x - {x^3}$ is a cubic polynomial.
View full question & answer→Question 451 Mark
Classify as linear, quadratic and cubic polynomials:
${x^2} + x{\text{ }}$
Answer${x^2} + x{\text{ }}$
We can observe that the degree of the polynomial ${x^2} + x{\text{ }}$ is $2.$
Therefore, we can conclude that the polynomial ${x^2} + x{\text{ }}$ is a quadratic polynomial.
View full question & answer→Question 461 Mark
Write the degree of the polynomial $: 3$
AnswerIt is a non-zero constant. So the degree of this polynomial is zero.
View full question & answer→Question 471 Mark
Write the degree of the polynomial: $5t - \sqrt{7}$
AnswerTerm with the highest power of $t = 5t$
Exponent of $t$ in this term $= 1$
$\therefore$ Degree of this polynomial $= 1$
View full question & answer→Question 481 Mark
Write the degree of the polynomial: $4-y^2$
AnswerTerm with the highest power of $\mathrm{y}=-\mathrm{y}^2$
Exponent of y in this term $=2$
$\therefore$ Degree of this polynomial $=2$.
View full question & answer→Question 491 Mark
Write the degree of the polynomial: $5 x^3+4 x^2+7 x$
AnswerIn the given polynomial, Term with the highest power of $x$ is $5 x^3$
Exponent of $x$ in this term $=3$
$\therefore$ Degree of the given polynomial is $3$ .
View full question & answer→Question 501 Mark
Give one example of a binomial of degree $35,$ and of a monomial of degree $100.$
AnswerThe binomial of degree $35$ can be $x^{35}+9$
The monomial of degree $100$ can be t${ }^{100}$
View full question & answer→Question 511 Mark
Write the coefficient of $x^2$ in $\sqrt{2} x-1$
AnswerSince $x^2$ is absent in given expression, therefore,
Coefficient of $x^2 = 0$
View full question & answer→Question 521 Mark
Write the coefficient of ${x^2}$ in $\frac{\pi }{2}{x^2} + x$
Answer$\frac{\pi }{2}{x^2} + x$
The coefficient of ${x^2}$ in the polynomial $\frac{\pi }{2}{x^2} + x$ is $\frac{\pi }{2}$.
View full question & answer→Question 531 Mark
Write the coefficient of $x^2$ in $2-x^2+x^3$
AnswerWe have,
Coefficient of $x^2=-1$
View full question & answer→Question 541 Mark
Write the coefficient of $x^2$ in $2+x^2+x$
Answer$2 + {x^2} + x$
The coefficient of ${x^2}$ in the polynomial $2 + {x^2} + x$ is $1.$
View full question & answer→Question 551 Mark
Is the expression ${x^{10}} + {y^3} + {t^{50}}$, polynomial in one variable or not$?$ State the reason for your answer.
Answer${x^{10}} + {y^3} + {t^{50}}$
We can observe that in the polynomial ${x^{10}} + {y^3} + {t^{50}}$, we have $x, y$ and $t$ as the variables and the powers of $x, y$ and $t$ in each term is a whole number.
Therefore, we conclude that ${x^{10}} + {y^3} + {t^{50}}$ is a polynomial but not a polynomial in one variable.
View full question & answer→Question 561 Mark
Is the expression $y + \frac{2}{y}$, polynomial in one variable or not$?$ State the reason for your answer.
Answer$y + \frac{2}{y}$
We can observe that in the polynomial $y + \frac{2}{y}$ ,we have $y$ as the only variable and the powers of $y$ in each term are not a whole number.
Therefore, we conclude that $y + \frac{2}{y}$ is not a polynomial in one variable.
View full question & answer→Question 571 Mark
Is the expression $3\sqrt t + t\sqrt 2$, polynomial in one variable or not? State the reason for your answer.
Answer$3\sqrt t + t\sqrt 2$
We can observe that in the polynomial $3\sqrt t + t\sqrt 2 $ we have t as the only variable and the
powers of t in each term are not a whole number.
Therefore, we conclude that $3\sqrt t + t\sqrt 2$ is not a polynomial in one variable.
View full question & answer→Question 581 Mark
Is the expression ${y^2} + \sqrt 2$, polynomial in one variable or not? State the reason for your answer.
Answer${y^2} + \sqrt 2$
We can observe that in the polynomial ${y^2} + \sqrt 2 $, we have y as the only variable and the powers of y in each term are a whole number.
Therefore, we conclude that ${y^2} + \sqrt 2$ is a polynomial in one variable.
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Is the expression $4{x^2} - 3x + 7{\text{ }}$, polynomial in one variable or not$?$ State the reason for your answer.
Answer$4{x^2} - 3x + 7{\text{ }}$
We can observe that in the polynomial $4{x^2} - 3x + 7{\text{ }}$
we have $x$ as the only variable and the powers of $x$ in each term are a whole number.
Therefore, we conclude that $4{x^2} - 3x + 7{\text{ }}$ is a polynomial in one variable.
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Find a zero of the polynomial $p(x) = 2x + 1$
AnswerFinding a zero of $p(x),$ is the same as solving the equation $p(x) = 0$
Now, $2x + 1 = 0$ gives us $x=-\frac{1}{2}$
So, $-\frac{1}{2}$ is a zero of the polynomial $2x + 1 $
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Check whether $–2$ and $2$ are zeroes of the polynomial $x + 2$
AnswerWe have, $p(x) = x + 2$
Then $p(2) = 2 + 2 = 4, p(–2) = –2 + 2 = 0$
Therefore, $–2$ is a zero of the polynomial $x + 2,$ but $2$ is not the zero of the polynomial.
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Evaluate $999^3$ using suitable identity.
AnswerWe have, $(999)^3=(1000-1)^3$
$=(1000)^3-(1)^3-3(1000)(1)(1000-1)$
$=1000000000-1-2997000$
$=997002999$
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Find the value of $p(t)=4 t^4+5 t^3-t^2+6$ at $t=a$.
Answer
We have, $p(t)=4 t^4+5 t^3-t^2+6$
On putting $t=a$ in $p(t)$, we get,
$p(a)=4(a)^4+5(a)^3-(a)^2+6$
$=4 a^4+5 a^3-a^2+6$
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Find the value of $p(x)=5 x^2-3 x+7$ at $x=1$
AnswerThe given polynomial is, $p(x)=5 x^2-3 x+7$
The value of the polynomial $p(x)$ at $x=1$ is given by $p(1)=5(1)^2-3(1)+7=5-3+7=9$
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Expand $(4 a-2 b-3 c)^2$
AnswerUsing Identity $(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x$,
we have $(4 a-2 b-3 c)^2=[4 a+(-2 b)+(-3 c)]^2$
$=(4 a)^2+(-2 b)^2+(-3 c)^2+2(4 a)(-2 b)+2(-2 b)(-3 c)+2(-3 c)(4 a)$
$=16 a^2+4 b^2+9 c^2-16 a b+12 b c-24 a c$
This is the required expansion.
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Write $(3 a+4 b+5 c)^2$ in expanded form.
AnswerComparing the given expression with $(x+y+z)^2$, we find that $x=3 a, y=4 b$ and $z=5 c$.
Therefore, using Identity $(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x$
we have $(3 a+4 b+5 c)^2=(3 a)^2+(4 b)^2+(5 c)^2+2(3 a)(4 b)+2(4 b)(5 c)+2(5 c)(3 a)$
$=9 a^2+16 b^2+25 c^2+24 a b+40 b c+30 a c$
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Factorise: $\frac{25}{4} x^{2}-\frac{y^{2}}{9}.$
Answer$a^2-b^2=(a+b)(a-b)$
$\frac{25 x^{2}}{4}-\frac{y^{2}}{9}=\left(\frac{5}{2} x\right)^{2}-\left(\frac{y}{3}\right)^{2}$
$=\left(\frac{5}{2} x+\frac{y}{3}\right)\left(\frac{5}{2} x-\frac{y}{3}\right)$
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Find the products using appropriate identities: $(x-3)(x+5)$
AnswerWe know the Identity i.e.. $(x+a)(x+b)=x^2+(a+b) x+a b$,
We have $(x-3)(x+5)=x^2+(-3+5) x+(-3)(5)=x^2+2 x-15$
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Find the products using appropriate identities: $(x+3)(x+3)$
AnswerWe have the Identity: $(x+y)^2=x^2+2 x y+y^2$.
Put $y=3$ in it,
we get $(x+3)(x+3)=(x+3)^2=x^2+2(x)(3)+(3)^2=x^2+6 x+9$
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Find the degree of the polynomial: $2$
AnswerThe only term here is $2$ which can be written as $2 x^0$. So the exponent of $x$ is $0$ . Hence, the degree of the polynomial is $0$ .
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Find the degree of the polynomial :
$2-y^2-y^3+2 y^8$
AnswerThe highest power of the variable is $8.$ Therefore, the degree of the polynomial is $8.$
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Find the degree of the polynomial :
$x^5-x^4+3$
AnswerThe highest power of the variable is $5.$ Therefore, the degree of the polynomial is $5.$
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Find the value of q(y) = 3y3 - 4y +$\sqrt{11}$at y = 2.
AnswerWe have, q(y) = 3y3 - 4y + $\sqrt{11}$
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 + $\sqrt{11}$
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Evaluate: (104)3 using a suitable identity
Answer(104)3 = (100 + 4)3
Using identity (x + y)3 = x3 + 3xy(x + y) + y3
We get,
(100 + 4)3 = (100)3 + 3 $\times$ 100 $\times$ 4 (100 + 4) + 43
= 10,00 000 + 1,200 $\times$ 104 + 64
= 10,00,000 + 1,24,800 + 64
= 11,24,864
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