Question 11 Mark
Factories : $64a^3 - 27b^3 - 144a^2b + 108ab^2$
Answer$64a^3 - 27b^3 - 144a^2b + 108ab^2$
$= (4a)^3 - (3b)^3 - 3(4a)(3b)(4a - 3b)$
$= (4a - 3b)^3$
$($Using Identity $(a – b)^3= a^3– b^3– 3ab (a – b))$
$= (4a - 3b)(4a - 3b)(4a- 3b)$
View full question & answer→Question 21 Mark
Factorise :$27 - 125a^3 - 135a + 225a^2$
Answer$27 - 125a^3 - 135a + 225a^2$
$= (3)^3 - (5a)^3 - 3(3)(5a)(3 - 5a)$
$= (3 - 5a)^3$
$($Using Identity $(a – b)^3= a^3– b^3– 3ab (a – b))$
$= (3 - 5a)(3 - 5a)(3 - 5a)$
View full question & answer→Question 31 Mark
Factorise : $8a^3 - b^3 - 12a^2b + 6ab^2$
Answer$8a^3 - b^3 - 12a^2b + 6ab^2$
$= (2a)^3 - (b)^3 + 3(2a)(b)(2a - b)$
$= (2a- b)^3$
$($Using Identity $(a – b)^3= a^3– b^3– 3ab (a – b))$
$= (2a - b)(2a - b)(2a - b)$
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 $(2a - 3b)^3$ in the expanded form
Answer$(2a - 3b)^3$
$= (2a)^3 - (3b)^3 - 3(2a)(3b)(2a - 3b)$
$($Using Identity $(a – b)^3= a^3– b^3– 3ab (a – b))$
$= 8a^3 - 27b^3 - 18ab(2a - 3b)$
$= 8a^3 - 27b^3 - 36a^2b + 54ab^2$
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 $(a + b + c)^2= a^2+ b^2+ c^2+ 2ab + 2bc + 2ca)$
$= \left(\frac{1}{4} a\right)^{2}+\left(-\frac{1}{2} b\right)^{2}$ + $+(1)^{2}+2\left(\frac{1}{4} a\right)\left(-\frac{1}{2} b\right)+2\left(-\frac{1}{2} b\right)(1)+2(1)\left(\frac{1}{4} a\right)$
$= \frac{1}{16} a^{2}+\frac{1}{4} b^{2}+1-\frac{1}{4} a b-b+\frac{1}{2} a$
View full question & answer→Question 81 Mark
Factorise $:4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz$
Answer$4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz$
$= (2x)^2 + (3y)^2 + (- 4z)^2 + 2(2x)(3y)+ 2(3y)(-4z) + 2(- 4z)(2x)$
$= {2x + 3y + (-4z)}^2 = (2x + 3y - 4z)^2$
$= (2x + 3y - 4z)(2x + 3y - 4z)$
View full question & answer→Question 91 Mark
Expand using appropriate identity: $(-2x + 5y - 3z)^2$
Answer$(-2x + 5y - 3z)^2$
$= {(-2x) + 5y + (-3z)}^2$
$= (-2x)^2+ (5y)^2 + (-3z)^2 + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)$
$= 4x^2 + 25y^2 + 9z^2 - 20xy - 30yz + 12zx$
View full question & answer→Question 101 Mark
Expand using appropriate identity: $(3a - 7b - c)^2$
Answer$(3a - 7b - c)^2= {3a + (- 7b) + (-c)}^2$
$= (3a)^2 + (-7b)^2 + (-c)^2 + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)$
$= 9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ca$
View full question & answer→Question 111 Mark
Expand using suitable identity: $(-2x + 3y + 2z)^2$
Answer$(- 2x + 3y + 2z)^2$
$= {(-2x) + 3y + (2z)}^2$
$= (-2x)^2 + (3y)^2 + (2z)^2 + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)$
$($Using Identity$ (a + b + c)^2= a^2+ b^2+ c^2+ 2ab + 2bc + 2ca)$
$= 4x^2 + 9y^2 + 4z^2 - 12xy+ 12yz - 8zx$
View full question & answer→Question 121 Mark
Expand using suitable identity: $(2x - y + z)^2$
Answer$(2x - y + z)^2$
$= {2x + (-y) + z}^2$
$= (2x)^2 + (-y)^2 + (z)^2+ 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)$
$($Using Identity $(a + b + c)^2= a^2+ b^2+ c^2+ 2ab + 2bc + 2ca)$
$= 4x^2 + y^2 + z^2 - 4xy - 2yz + 4zx$
View full question & answer→Question 131 Mark
Expand using suitable identity: $(x + 2y + 4z)^2$
Answer$(x + 2y + 4z)^2$
$= (x)^2 + (2y)^2 + (4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)$
$($Using Identity $(a + b + c)^2= a^2+ b^2+ c^2+ 2ab + 2bc + 2ca)$
$= x^2 + 4y^2+ 16z^2 + 4xy + 16yz + 8zx$
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 $(x+a)(x+b)$
$=x^2 + (a+b)x + ab$
$(100+3)(100+7)$
$=(100)^2 + (3 + 7)(100) + 3\times 7$
$=10000 + 1000 + 21$
$= 11021$
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 $x^3 + y^3 + z^3 = 3xyz.$
AnswerWe know that
$x^3+ y^3 + z^3 - 3xyz$
$ = (x+ y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$
$($Using Identity $a^3+ b^3+ c^3– 3abc= (a + b + c)(a^2+ b^2+ c^2– ab – bc – ca))$
$= (0) (x^2 + y^2 + z^2 - xy - yz - zx) (\because x + y + z = 0)$
$= 0$
$\Rightarrow x^3 + y^3+ z^3$
$ = 3xyz.$
View full question & answer→Question 221 Mark
Use suitable identity to find the product: $(3x + 4)(3x - 5)$
Answer$(3x + 4)(3x - 5)$
$= (3x + 4){3x + (-5)}$
Using identity $(x + a)(x + b) = x^2+ (a + b) x + ab$
$= (3x)^2 + {4 + (-5)}(3x) + (4)(-5)$
$= 9x^2 - 3x - 20$
View full question & answer→Question 231 Mark
Factorise : $27x^3 + y^3 + z^3 - 9xyz.$
Answer$27x^3 + y^3 + z^3 - 9xyz.$
$(3x)^3 + (y)^3 + (z)^3 - 3(3x)(y)(z)$
$= (3x + y + z){(3x)^2 + (y)^2 + (z)^2 - (3x)(y) - (y)(z) - (z)(3x)}$
$($Using Identity $a^3+ b^3+ c^3– 3abc= (a + b + c)(a^2+ b^2+ c^2– ab – bc – ca))$
$= (3x + y + z)(9x^2 + y^2 + z^2 - 3xy - yz - 3zx).$
View full question & answer→Question 241 Mark
Factorise :$64m^3 - 343n^3$
Answer$64m^3 - 343n^3$
$= (4m)^3 - (7n)^3 $
$= (4m - 7n){(4m)^2 + (4m)(7n) + (7n)^2}$
$= (4m - 7n)(16m^2 + 28mn + 49n^2)$
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 $x^3 + 3x^2 + 3x + 1$ is divided by $5 + 2x.$
AnswerLet $p(x) = x^3 + 3x^2 + 3x + 1$
$5 + 2x = 0$
$\Rightarrow2x = -5$
$\Rightarrow x = -\frac{5}{2}$
$= (-\frac{5}{2})^3 + 3(-\frac{5}{2})^2 + 3(-\frac{5}{2}) + 1$
$= -\frac{125}{8} + \frac{75}{4} - \frac{15}{2} + 1$
$= -\frac{27}{8}$
View full question & answer→Question 271 Mark
Find the remainder when $x^3 + 3x^2 + 3x + 1$ is divided by $x + \pi$
AnswerLet $p(x) = x^3 + 3x^2 + 3x + 1$
$x + \pi = 0$
$\Rightarrow x = -\pi$
$\therefore$ 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
Answerx + 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
Answerp(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
Answerp(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 + 5
p(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 $5x – 4x^2 + 3$ at $x = 2$
AnswerLet $f(x) = 5x – 4x^2 + 3$
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 $5x – 4x^2 + 3$ at $x = -1$
AnswerLet $f(x) = 5x \ – 4x^2 + 3$
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 $5x - 4x^2 + 3$ at $x = 0$
AnswerLet $f(x) = 5x - 4x^2 + 3$
We need to substitute $x = 0$ in the polynomial $f(x) = 5x - 4x^2 + 3$ 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 $5x - 4x^2 + 3$ 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
Answer3t
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 $y = \ – y^2$
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: $5x^3 + 4x^2 + 7x$
AnswerIn the given polynomial, Term with the highest power of $x$ is $5x^3$
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→