Chapter 2 : Polynomials

Exercise – 1

Q.1.Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer:

(i)$$4x^{2}-3x+7$$

Answer: It is a polynomial in one variable.

(ii) $$y^{2}+\sqrt{2}$$

Answer: It is a polynomial in one variable.

(iii) $$3\sqrt{t}+t\sqrt{2}$$

Answer: It is not a polynomial since the power of the variable is not a whole number.

(iv) $$y+\frac{2}{y}$$

Answer: It is not a polynomial since the power of the variable is not a whole number.

(v) $$x^{10}+y^{3}+t^{50}$$

Answer: It is a polynomial in three variables.

Q.2. Write the coefficients of a2 in each of the following:

(i) $$2 + a^{2} + a$$

Answer: Coefficient of $$a^{2}$$ is 1.

(ii) $$2 – a^{2} + a^{3}$$

Answer: Coefficient of $$a^{2}$$ is -1.

(iii) $$\frac{\pi }{2}x^{2}+x$$

Answer: Coefficient of $$a^{2}$$ is $$\frac{\pi }{2}$$.

(iv)
$$\sqrt{2}x-1$$

Answer: Coefficient of $$a^{2}$$ is 0.

Q.3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Answer: $$3x^{35}+5\; and\; 4x^{100}$$

Q.4. Write the degree of each of the following polynomials:

(i) $$5a^{3} + 4a^{2} + 7a$$

(ii) $$3-b^{2}$$

(iii) $$5t – \sqrt{8}$$

(iv)8

Q.5. Classify the following as linear, quadratic and cubic polynomial.

(i) $$a^{2} + a$$

(ii)$$a-a^{3}$$

(iii) $$y + y^{2} +4$$

(iv)1 + x

(v) 3a

(vi) $$a^{2}$$

(vii) $$6a^{3}$$

Exercise – 2

Q.1. Find the value of the polynomial at$$f(x)= 5a -4a^{2}+ 3$$ at

(i) a= 0

(ii) a = – 1

(iii) a = 2

Let $$f(x)= 5a -4a^{2}+ 3$$

(i) When a=0

$$f(0) = 5(0) + 4(0)^{2} + 3= 3$$

(ii) When a=-1

$$f(a) = 5a + 4a^{2}+ 3\\ f(-1) = 5(-1) + 4(-1)^{2} + 3 = -5 – 4 + 3 = -6$$

(iii) When a=2

$$f(a) = 5a + 4a^{2} + 3\\ f(2) = 5(2) + 4(2)^{2} + 3 = 10 – 16 + 3 = -3$$

Q.2. Find p(0), p(1) and p(2) for each of the following polynomials:

(i) $$p(y) = y^{2}- y + 1$$

Answer: $$p(y) = y2 – y + 1\\ ∴ p(0) = (0)^{2}- (0) + 1 = 1\\ p(1) = (1)^{2} – (1) + 1 = 1\\ p(2) = (2)^{2} – (2) + 1 = 3$$

(ii) $$p(a) = 2 + a + 2a^{2}- a^{3}$$

Answer: $$p(a) = 2 +a + 2a^{2}-a^{3}\\ ∴ p(0) = 2 + 0 + 2 (0)^{2} – (0)^{3} = 2\\ p(1) = 2 + 1+ 2(1)^{2} – (1)^{3} = 2 + 1 + 2 – 1 = 4\\ p(2) = 2 + 2 + 2(2)^{2} – (2)^{3} = 2 + 2 + 8 – 8 = 4$$

(iii) $$p(x) = x^{3}$$

Answer: $$p(x) = x3\\ ∴ p(0) = (0)^{3}= 0\\ p(1) = (1)^{3} = 1\\ p(2) = (2)^{3} = 8$$

(iv) $$p(a) = (a- 1) (a+ 1)$$

Answer: $$p(a) = (a – 1) (a + 1)\\ ∴ p(0) = (0 – 1) (0 + 1) = (- 1) (1) = – 1\\ p(1) = (1 – 1) (1 + 1) = 0 (2) = 0\\ p(2) = (2 – 1 ) (2 + 1) = 1(3) = 3$$

Q.3. Verify whether the following are zeroes of the polynomial, indicated against them.

(i) $$p(x) = 3x + 1, x = -\frac{1}{3}$$

Answer: For, $$x = -\frac{1}{3}$$

$$p(x) = 3x + 1\\ ∴ p(-\frac{1}{3}) = 3(-\frac{1}{3}) + 1=-1+1=0$$

$$∴ -\frac{1}{3}$$ is a zero of p(x).

(ii) $$p(x) = 5x – \pi , x = \frac{4}{5}$$

Answer: For, $$x = \frac{4}{5}$$

$$p(x) = 5x – \pi \\ ∴ p(\frac{4}{5}) = 5(\frac{4}{5}) – \pi=4-\pi$$

$$∴ \frac{4}{5}$$ is not a zero of p(x).

(iii) $$p(x) = x^{2}- 1, x = 1, -1$$

Answer: For, $$x = 1,-1$$

$$p(x) = x^{2}- 1\\ ∴ p(1) = x^{1}- 1=1-1=0\\ p(-1) = x^{-1}- 1=1-1=0$$

$$∴ 1,-1$$ are zeros of p(x).

(iv) $$p(x) = (x + 1) (x – 2), x = -1, 2$$

Answer: For, $$x = -1,2$$

$$p(x) = (x + 1) (x – 2)\\ ∴ p(-1) = (-1 + 1) (-1 – 2)\\ =((0)(-3))\\ =0\\ \\ p(2) = (2 + 1) (2 – 2)\\ =(3)(0)=0$$

$$∴ -1,2$$ are zeros of p(x).

(v) $$p(x) = x^{2}, x = 0$$

Answer: For, $$x = 0$$

$$p(x) = 0^{2}=0$$

$$∴ 0$$ is a  zero of p(x).

(vi) $$p(x)=lx+m,x=-\frac{m}{l}$$

Answer: For, $$x=-\frac{m}{l}$$

$$p(x)=lx+m\\ ∴ p(-\frac{m}{l} )=l(-\frac{m}{l} )+m=-m+m=0$$

$$∴-\frac{m}{l}$$ is a zero of p(x).

(vii) $$p(x)=3x^{2}-1,x=-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}$$

Answer: For, $$x=-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}$$
$$p(x)=3x^{2}-1\\ ∴ p(-\frac{1}{\sqrt{3}})=3(-\frac{1}{\sqrt{3}})^{2}-1\\ =3(\frac{1}{3})-1=1-1=0\\ \\ ∴ p(\frac{2}{\sqrt{3}})=3(\frac{2}{\sqrt{3}})^{2}-1\\ =3(\frac{4}{3})-1=4-1=3\neq 0$$

$$∴ -\frac{1}{\sqrt{3}}$$ is a zero of p(x) but $$\frac{2}{\sqrt{3}}$$ is not a zero of p(x).

(viii) $$p(x) = 2x + 1, x = \frac{1}{2}$$
Answer: For, $$x=\frac{1}{2}$$

$$p(x) = 2x + 1\\ ∴ p(\frac{1}{2} ) = 2(\frac{1}{2}) + 1=1+1=2\neq 0$$

$$∴\frac{1}{2}$$ is not a  zero of p(x).

Q.4. Find the zero of the polynomial in each of the following cases:

(i)p(x) = x + 5

Answer: $$p(x) = x + 5 \\ \Rightarrow x + 5=0 \Rightarrow x=-5$$

$$∴$$-5 is a zero polynomial of the polinomila p(x).

(ii) p(x) = x – 5

Answer: $$p(x) = x -5 \\ \Rightarrow x -5=0 \Rightarrow x=5$$

$$∴$$5 is a zero polynomial of the polinomila p(x).

(iii) p(x) = 2x + 5

Answer: $$p(x) =2 x + 5 \\ \Rightarrow 2x + 5=0 \Rightarrow 2x=-5\\ \Rightarrow x=\frac{-5}{2}$$

$$∴$$$$x=\frac{-5}{2}$$ is a zero polynomial of the polynomial p(x).

(iv)p(x) = 3x – 2

Answer: $$p(x) =3 x – 2 \\ \Rightarrow 3x -2=0 \Rightarrow 3x=2\\ \Rightarrow x=\frac{2}{3}$$

$$∴$$$$x=\frac{2}{3}$$ is a zero polynomial of the polynomial p(x).

(v)p(x) = 3x

Answer: $$p(x) =3 x \\ \Rightarrow 3x =0 \Rightarrow x=0$$

$$∴$$0 is a zero polynomial of the polynomial p(x).

Exercise – 3

Q.1.Find the remainder when $$x^{3}+3x^{2}+3x+1$$ is divided by

(i) x+1

$$x+1=0\\ \Rightarrow x=-1$$ $$∴\;Remainder=p(-1)=(-1)^{3}+3(-1)^{2}+3(-1)+1=-1+3-3+1=0$$

(ii) $$x-\frac{1}{2}$$

$$x-\frac{1}{2}=0\\ \Rightarrow x=\frac{1}{2}$$ $$∴\;Remainder (\frac{1}{2})^{3}+3(\frac{1}{2})^{2}+3(\frac{1}{2})+1\\ =\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1=\frac{27}{8}$$

(iii) x

$$∴\;Remainder= (0)^{3}+3(0)^{2}+3(0)+1\\ =1$$

(iv) $$x+\pi$$

$$x+\pi=0\\ \Rightarrow x=-\pi$$ $$∴ \;Remainder= (-\pi)^{3}+3(-\pi)^{2}+3(-\pi)+1=-\pi ^{3}+3\pi ^{2}-3\pi +1$$

(v) 5+2x

$$5+2x=0\\ \Rightarrow 2x=-5\Rightarrow x=-\frac{5}{2}$$ $$∴\; Remainder=(-\frac{5}{2})^{3}+3(-\frac{5}{2})^{2}+3(-\frac{5}{2})+1=-\frac{125}{8}+\frac{75}{4}-\frac{15}{2}+1=-\frac{27}{8}$$

Q.2.Find the remainder when $$x^{3}-ax^{2}+6x-a$$ is divided by x-a.

Let $$p(x)=x^{3}-ax^{2}+6x-a\\ x-a=0$$

$$∴ x=a$$ $$Remainder=(a)^{2}-a(a^{2})+6(a)-a\\ =a^{3}-a^{3}+6a-a\\ =5a$$

Q.3.Check whether 7+3x is a factor of $$3x^{3}+7x$$.

$$7+3x=0\Rightarrow 3x=-7$$ only if 7+3x divides $$3x^{3}+7x$$ leaving no reaminder.

Let $$p(x)=3x^{3}+7x$$

$$7+3x=0\Rightarrow 3x=-7\Rightarrow x=-\frac{7}{3}$$ $$∴ Remainder=3(-\frac{7}{3})^{3}+7(-\frac{7}{3})=-\frac{343}{9}-\frac{49}{3}=-\frac{490}{9}\neq 0$$

$$∴$$7+3x is not a factor of $$3x^{3}+7x$$

Exercise – 4

Q.1. Determine which of the following polynomials has (x + 1) a factor:

(i)
$$x^{3} + x^{2}+ x + 1$$

Let p(x)= $$x^{3} + x^{2}+ x + 1$$

The zero of x+1 is -1.

$$p(-1)=(-1)^{3} + (-1)^{2}+ (-1) + 1\\ =-1+1-1+1=0$$

$$∴$$By factor theorem, x+1 is a factor of $$x^{3} + x^{2}+ x + 1$$

(ii) x4 + x3 + x2 + x + 1

Let p(x)= $$x^{4} + x^{3}+x^{2} +x + 1$$

The zero of x+1 is -1.

$$p(-1)=(-1)^{4} + (-1)^{3}+ (-1)^{2} + (-1)+1\\ =1-1+1-1+1=1\neq 0$$ $$∴$$By factor theorem, x+1 is a factor of $$x^{4} + x^{3}+x^{2} +x + 1$$

(iii) x4 + 3x3 + 3x2 + x + 1

Let p(x)= $$x^{4} +3 x^{3}+3x^{2} +x + 1$$

The zero of x+1 is -1.

$$p(-1)=(-1)^{4} + 3(-1)^{3}+ 3(-1)^{2} + (-1)+1\\ =1-3+3-1+1=1\neq 0$$ $$∴$$By factor theorem, x+1 is a factor of $$x^{4} + 3x^{3}+3x^{2} +x + 1$$

(iv) $$x^{3} – x^{2} – (2 + \sqrt{2})x + \sqrt{2}$$

Let p(x)= $$x^{3} – x^{2} – (2 + \sqrt{2})x + \sqrt{2}$$

The zero of x+1 is -1.

$$p(-1)=(-1)^{3} – (-1)^{2} – (2 + \sqrt{2})(-1) + \sqrt{2}$$

$$∴$$By factor theorem, x+1 is not a factor of $$x^{3} – x^{2} – (2 + \sqrt{2})x + \sqrt{2}$$

Q.2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

(i)
$$p(x) = 2x^{3} + x^{2} – 2x – 1$$, g(x) = x + 1

Answer: $$p(x) = 2x^{3} + x^{2} – 2x – 1$$, g(x) = x + 1

g(x)=0

$$\Rightarrow x+1=0\Rightarrow x=-1$$

$$∴$$Zero of g(x) is -1.

Now, $$p(-1) = 2(-1)^{3} + (-1)^{2} – 2(-1) – 1\\ =-2+1+2-1=0$$

$$∴$$By factor theorem, g(x) is a factor of p(x).
(ii) $$p(x) = x^{3}+ 3x^{2} + 3x + 1$$, g(x) = x + 2

Answer: $$p(x) = x^{3} + 3x^{2} +3x + 1$$, g(x) = x + 2

g(x)=0

$$\Rightarrow x+2=0\Rightarrow x=-2$$

$$∴$$Zero of g(x) is -2.

Now, $$p(-2) = (-2)^{3} + 3(-2)^{2} +3(-2) + 1\\ =-8+12-6+1=-1\neq=0$$

$$∴$$By factor theorem, g(x) is not a factor of p(x).
(iii) $$p(x) = x^{3} – 4 x^{2} + x + 6$$, g(x) = x – 3

Answer: $$p(x) = x^{3} – 4x^{2} +x + 6$$, g(x) = x -3

g(x)=0

$$\Rightarrow x-3=0\Rightarrow x=3$$

$$∴$$Zero of g(x) is 3.

Now, $$p(3) = (3)^{3} -4(3)^{2} +(3) + 6\\ =27-36+3+6=0$$

$$∴$$By factor theorem, g(x) is a factor of p(x).

Q.3. Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

(i)$$p(x) = x^{2}+ x + k$$

Answer: If x-1 is a factor of p(x), then p(1)=0   |By Factor Theorem $$\Rightarrow (1)^{2}+ (1) + k=0\Rightarrow 1+1+k=0$$

$$\Rightarrow2+k=0\Rightarrow k=-2$$

(ii) $$p(x) = 2x^{2} + kx + \sqrt{2}$$

Answer: If x-1 is a factor of p(x), then p(1)=0

$$\Rightarrow 2(1)^{2}+ k(1) +\sqrt{2} =0\Rightarrow 2+k+\sqrt{2}=0\Rightarrow k=-(2+\sqrt{2})$$

(iii) $$p(x) = kx^{2} – \sqrt{2}x + 1$$

Answer: If x-1 is a factor of p(x), then p(1)=0

By Factor Theorem

$$\Rightarrow k(1)^{2}-\sqrt{2}(1)+1=0\\ \Rightarrow k=\sqrt{2}-1$$
(iv) $$p(x) = kx^{2} – 3x + k$$

Answer: If x-1 is a factor of p(x), then p(1)=0

By Factor Theorem

$$\Rightarrow k(1)^{2} – 3(1) + k=0\Rightarrow k-3+k=0\\ \Rightarrow 2k-3=0\\ \Rightarrow k=\frac{3}{2}$$

Q.4. Factorize:

(i) $$12x^{2} – 7x + 1$$

$$12x^{2} – 7x + 1=12x^{2} – 4x-3x + 1\\ =4x(3x-1)-1(3x-1) =(4x-1)(3x-1)$$

Let p(x)= $$12x^{2} – 7x + 1$$

Then p(x)= $$12(x^{2} – \frac{7}{12}x + \frac{1}{12})=12q(x)$$

Where q(x)= $$x^{2} – \frac{7}{12}x + \frac{1}{12}$$

By trial, we find that

$$q(\frac{1}{3})=(\frac{1}{3})^{2} – \frac{7}{12}(\frac{1}{3}) + \frac{1}{12}\\ =\frac{4-7+3}{36}=\frac{0}{36} =0$$

$$∴$$By Factor Theorem,

$$(x-\frac{1}{3})$$ is a factor of q(x).

Similarly, by trial, we find that

$$q(\frac{1}{4})=(\frac{1}{4})^{2} – \frac{7}{12}(\frac{1}{4}) + \frac{1}{12}\\ =\frac{3-7+4}{18}=\frac{0}{48} =0$$

$$∴$$By Factor Theorem,

$$(x-\frac{1}{4})$$ is a factor of q(x).

Therefore, $$12x^{2}-7x+1=12(x-\frac{1}{3})(x-\frac{1}{4})\\ =12(\frac{3x-1}{3})(\frac{4x-1}{4})=(3x-1)(4x-1)$$

(ii)$$2x^{2} + 7x + 3$$

$$2x^{2} + 7x + 3=2x^{2} + 6x+x + 3\\ =2x(x+3) +1 (x + 3)\\ =(x+3)(2x+1)$$

Let p(x)= $$2x^{2} + 7x + 3$$

Then p(x)= $$2(x^{2} + \frac{7}2{}x + \frac{3}{2})=2q(x)$$
Where q(x)= $$x^{2} + \frac{7}2{}x + \frac{3}{2}$$

By trial, we find that

$$q(-3)=(-3)^{2} – \frac{7}{2}(-3) + \frac{3}{2}\\ =9-\frac{21}{2}+\frac{3}{2}=0$$

$$∴$$By Factor Theorem,

$$(x-3),i.e.(x+3)$$ is a factor of q(x).

Similarly, by trial, we find that

$$q(-\frac{1}{2})=(-\frac{1}{2})^{2} +\frac{7}{12}(-\frac{1}{2}) + \frac{3}{2}\\ =\frac{1}{3}-\frac{7}{4}+\frac{3}{4} =0$$

$$∴$$By Factor Theorem,

$$(x-(-\frac{1}{2})),i.e.(x+\frac{1}{2}) \;is\; a\; factor\; of\; q(x).\\ Therefore, 2x^{2}+7x+3=2(x+3)(x+\frac{1}{2})\\ =2(x+3)(\frac{2x+1}{2})\\ =(x+3)(2x+1)$$

By Simpler method,

(iii) $$6x^{2}+5x-6 =6x^{2}+ 9x – 4x – 6 =3x (2x + 3) – 2 (2x + 3) = (2x + 3) (3x – 2)$$

(iv)3x2 – x – 4 =$$3x^{2} – x – 4 \\ =3×2 – 4x + 3x – 4 \\ = x (3x – 4) + 1 (3x – 4)\\ = (3x – 4) (x + 1)$$

Q.5. Factorize:

(i)
$$x^{3} – 2x^{2} – x + 2$$

Answer: Let $$p(x) = x^{3} – 2x^{2} – x + 2$$
Factors of 2 are ±1 and ± 2
By trial method, we find that
p(1) = 0
So, (x+1) is factor of p(x)
Now,
$$p(x) = x^{3} – 2x^{2} – x + 2$$

$$p(-1) = (-1)^{3} – 2(-1)^{2} – (-1) + 2=-1-1+1+2=0$$

Therefore, (x+1) is the factor of  p(x)

Now, Dividend = Divisor × Quotient + Remainder

$$(x+1) (x^{2} – 3x + 2)\\ = (x+1) (x^{2} – x – 2x + 2)\\ = (x+1) (x(x-1) -2(x-1))\\ = (x+1) (x-1) (x+2)$$

(ii) $$x^{3} – 3x^{2} – 9x – 5$$
Answer: Let p(x) = $$x^{3} – 3x^{2} – 9x – 5$$
Factors of 5 are ±1 and ±5
By trial method, we find that
p(5) = 0
So, (x-5) is factor of p(x)
Now,
p(x) = $$x^{3} – 3x^{2} – 9x – 5$$

p(5) = $$(5)^{3} – 3(5)^{2} – 9(5) – 5=125-75-45-5=0$$

Therefore, (x-5) is the factor of  p(x)

Now, Dividend = Divisor × Quotient + Remainder

$$(x-5) (x^{2} + 2x + 1)\\ = (x-5) (x^{2} + x + x + 1)\\ = (x-5) (x(x+1) +1(x+1))\\ = (x-5) (x+1) (x+1)$$

(iii) $$x^{3} + 13x^{2} + 32x + 20$$

Answer: Let p(x) = $$x^{3} + 13x^{2} + 32x + 20$$

Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20
By trial method, we find that
p(-1) = 0
So, (x+1) is factor of p(x)
Now,
p(x) = $$x^{3} + 13x^{2} + 32x + 20$$

p(-1) = $$(-1)^{3} + 13(-1)^{2} + 32(-1) + 20=-1+13-32+20=0$$

Therefore, (x+1) is the factor of  p(x)

Now, Dividend = Divisor × Quotient + Remainder

$$(x+1) (x^{2} + 12x + 20)\\ = (x+1) (x^{2} + 2x + 10x + 20)\\ = (x-5) {x(x+2) +10(x+2)}\\ = (x-5) (x+2) (x+10)$$

(iv) $$2y^{3} + y^{2} – 2y – 1$$
Answer: Let p(y) = $$2y^{3} + y^{2} – 2y – 1$$

Factors of ab = $$2\times (-1)$$= -2 are ±1 and ±2
By trial method, we find that
p(1) = 0
So, (y-1) is factor of p(y)
Now,
p(y) = $$2y^{3} + y^{2} – 2y – 1$$

p(1) = $$2(1)^{3} + (1)^{2} – 2(1) – 1=2+1-2=0$$

Therefore, (y-1) is the factor of  p(y)

Now, Dividend = Divisor × Quotient + Remainder

$$(y-1) (2y^{2} + 3y + 1)\\ = (y-1) (2y^{2} + 2y + y + 1)\\ = (y-1) (2y(y+1) +1(y+1))\\ = (y-1) (2y+1) (y+1)$$

Exercise – 5

Q.1.Use suitable identities to find the following products:

(i)(x + 4) (x + 10)

Answer: $$(x + 4) (x + 10) = x^{2} + (4 + 10)x + (4 \times 10)\\ = x^{2} + 14x + 40$$
(ii)(x + 8) (x – 10)

Answer: $$(x + 8) (x -10 ) = x^{2} + (8+(- 10))x + (8\times(-10))\\ = x^{2} +(8-10)x – 80\\ =x^{2}-2x-80$$

(iii)(3x + 4) (3x – 5)

Answer: $$(3x + 4)(3x -5) = (3x) ^{2} + {4 + (-5)}3x + {4×(-5)}\\ = 9x^{2} + 3x(4 – 5) – 20\\ = 9x^{2} – 3x – 20$$
(iv)$$(y^{2} + \frac{3}{2}) (y^{2} – \frac{3}{2})$$

Answer: $$(y^{2} + \frac{3}{2}) (y^{2} – \frac{3}{2}) = (y^{2})^{2} – (\frac{3}{2})^{2}\\ = y^{4} – \frac{9}{4}$$

Q.2. Evaluate the following products without multiplying directly:

(i)103 × 107

Answer:   $$103 \times 107 = (100 + 3)\times (100 + 7)\\ = (100)^{2} + (3 + 7)(100 + (3 \times 7))\\ = 10000 + 1000 + 21 \\ = 11021$$

(ii) 95 × 96

Answer: $$95 \times 96 = (90 + 5) \times (90 + 6)\\ = (90)^{2} + 90(5 + 6) + (5 \times 6) \\ = 8100 + 990 + 30 = 9120$$

(iii) 104 × 96
Answer: $$104 \times 96 = (100 + 4) \times (100 – 4)\\ = (100)^{2} – (4)^{2}\\ = 10000 – 16 \\ = 9984$$

Q.3. Factorize the following using appropriate identities:

(i)$$9x^{2} + 6xy + y^{2}$$

Answer: $$9x^{2} + 6xy + y^{2} \\ = (3x) ^{2} + (2\times 3x\times y) + y^{2}\\ = (3x + y)^{2} \\ = (3x + y) (3x + y)$$

(ii) $$4y^{2}- 4y + 1$$

Answer: $$4y^{2}- 4y + 1 \\ =(2y)^{2} – (2\times 2y\times 1) + 1^{2}\\ = (2y – 1)^{2}\\ = (2y – 1) (2y – 1)$$

(iii) $$x2 – \frac{y^{2}}{100}$$

Answer: $$x^{2} – (\frac{y}{10})^{2} = (x – \frac{y}{10}) (x + \frac{y}{10})$$

Q.4. Expand each of the following, using suitable identities:

(i) $$(x + 2y + 4z)^{2}$$

(ii) $$(2x- y + z)^{2}$$

(iii) $$(-2x+3y+2z)^{2}$$

(iv) $$[\frac{1 }{4}a – \frac{1}{2} b + 1]^{2}$$

(i) $$(x + 2y + 4z)^{2}$$

Using identity,

$$(x + 2y + 4z)^{2} \\ = x^{2} + (2y)^{2} + (4z)^{2} + (2\times x\times 2y) + (2\times 2y\times 4z) + (2\times 4z\times x)\\ = x^{2} + 4y^{2} + 16z2 + 4xy + 16yz + 8xz$$
(ii)  $$(2x- y + z)^{2}$$
Using identity,

$$(2x-y + z)^{2}\\ = (2x)^{2} + (-y)^{2}+ z^{2} + (2\times 2x\times -y) + (2\times -y\times z) + (2\times z\times 2x) \\ = 4x^{2} + y^{2} + z^{2} – 4xy – 2yz + 4xz$$

(iii) $$(-2x+3y+2z)^{2}$$

Using identity,

$$(-2x+ 3y + 2z)^{2} \\ =(-2x)^{2} + (3y)^{2} + (2z)^{2} + (2\times -2x\times 3y) + (2\times 3y\times 2z) + (2\times 2z\times -2x)\\ = 4x^{2} + 9y^{2} + 4z^{2} – 12xy + 12yz – 8xz$$
(iv) $$[\frac{1 }{4}a – \frac{1}{2} b + 1]^{2}$$
Using identity,
$$[\frac{1}{4} a -\frac{ 1}{2} b + 1]^{2} = (\frac{1}{4} a)^{2} + (-\frac{1}{2} b)^{2} + 1^{2} + (2\times \frac{1}{4} a\times -\frac{1}{2} b) + (2\times -\frac{1}{2} b\times 1) + (2\times 1\times \frac{1}{4} a) \\ = \frac{1}{16} a^{2} + \frac{1}{4} b^{2} + 1 – \frac{1}{4} ab – b + \frac{1}{2} a$$

Q.5. Factorize:
(i) $$4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz$$

(ii) $$2x^{2} + y^{2} + 8z^{2} – 2\sqrt{2} xy + 4\sqrt{2} yz – 8xz$$

(i) $$4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz$$

$$4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz$$

$$= (2x)^{2} + (3y)^{2} + (-4z)^{2} + (2\times 2x\times 3y) + (2\times 3y\times -4z) + (2\times -4z\times 2x)\\ = (2x + 3y – 4z)^{2}\\ = (2x + 3y – 4z) (2x + 3y – 4z)$$
(ii) $$2x^{2} + y^{2} + 8z^{2} – 2\sqrt{2} xy + 4\sqrt{2} yz – 8xz$$

$$2x^{2} + y^{2} + 8z^{2} – 2\sqrt{2} xy + 4\sqrt{2} yz – 8xz$$
$$= (-\sqrt{2}x)^{2} + (y)^{2} + (2\sqrt{2}z)^{2} + (2\times -\sqrt{2}x\times y) + (2\times y\times 2\sqrt{2}z) + (2\times 2\sqrt{2}z\times -\sqrt{2}x)\\ = (-\sqrt{2}x + y + 2\sqrt{2}z)^{2}\\ = (-\sqrt{2}x + y + 2\sqrt{2}z) (-\sqrt{2}x + y + 2\sqrt{2}z)$$

Q.6. Write the following cubes in expanded form:

(i)$$(2x + 1)^{3}$$

(ii) $$(2a -3b)^{3}$$

(iii) $$[\frac{3}{2} x + 1]^{3}$$

(iv) $$[ x -\frac{2}{3} y]^{3}$$

(i)$$(2x + 1)^{3}$$

$$(2x + 1)^{3} = (2x)^{3} + 1^{3} + (3\times 2x\times 1)(2x + 1)\\ = 8x^{3} + 1 + 6x(2x + 1)\\ = 8x^{3} + 12x^{3} + 6x + 1$$

(ii) $$(2a -3b)^{3}$$

$$(2a-3b)^{3}=(2a)^{3}-(3b)^{3} – (3\times 2a\times 3b)(2a – 3b)\\ = 8a^{3} – 27b^{3} – 18ab(2a – 3b)\\ = 8a^{3} – 27b^{3} – 36a^{2}b + 54ab^{2}$$

(iii) $$[\frac{3}{2} x + 1]^{3}$$

$$[\frac{3}{2} x + 1]^{3} = (\frac{3}{2} x)^{3} + 1^{3} + (3\times\frac{ 3}{2} x\times 1)(\frac{3}{2} x + 1)\\ = \frac{27}{8} x^{3}+ 1 + \frac{9}{2}x (\frac{3}{2} x + 1)\\ = \frac{27}{8} x3 + 1 + \frac{27}{4}^{2} + \frac{9}{2} x\\ = \frac{27}{8} x3 + \frac{27}{4} ^{2} + \frac{9}{2} x + 1$$
(iv) $$[ x -\frac{2}{3} y]^{3}$$
$$[x – \frac{2}{3} y]^{3} = (x)^{3} – (\frac{2}{3} y)^{3} – (3\times x\times \frac{2}{3} y)(x – \frac{2}{3} y)\\ = (x)^{3} – \frac{8}{27}y^{3} – 2xy(x – \frac{2}{3} y)\\ = (x)^{3} – \frac{8}{27}y3 – 2x^{2}y + \frac{4}{3}xy^{2}$$

Q.7. Evaluate the following using suitable identities:

(i)$$(99)^{3}$$

(ii)$$(102)^{3}$$

(iii) $$(998)^{3}$$

(i) $$(99)^{3} = (100 – 1)^{3}$$

$$(100 – 1)^{3} = (100)^{3} – 1^{3} – (3\times 100\times 1)(100 – 1)$$

= 1000000 – 1 – 300(100 – 1)

= 1000000 – 1 – 30000 + 300 = 970299

(ii) $$(102)^{3} = (100 + 2)^{3}$$
$$(100 + 2)^{3} = (100)^{3} + 2^{3}+ (3\times 100\times 2)(100 + 2)$$
= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200= 1061208

(iii) $$(998)^3= (1000 – 2)^{3}$$

$$= (1000)^{3} – 2^{3} – (3\times 1000\times 2)(1000 – 2)$$

= 1000000000 – 8 – 6000(1000 – 2)

= 1000000000 – 8- 6000000 + 12000 = 994011992

Q.8. Factorise each of the following:
(i)
$$8a^{3}+ b^{3} + 12a^{2}b + 6ab^{2}$$

(ii) $$8a^{3} – b^{3} – 12a^{2}b + 6ab^{2}$$

(iii)27 – 125a3 – 135a + 225a2

(iv) $$64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}$$

(v) $$27p^{3} – \frac{1}{216}- \frac{9}{2} p^{2} + \frac{1}{4} p$$

(i) $$8a^{3}+ b^{3} + 12a^{2}b + 6ab^{2}$$

$$8a^{3} + b^{3}+ 12a^{2}b + 6ab^{2}\\ = (2a)^{3} + b^{3} + 3(2a)^{2}b + 3(2a)(b)^{2}\\ = (2a + b)^{3}\\ = (2a + b)(2a + b)(2a + b)$$
(ii) $$8a^{3} – b^{3} – 12a^{2}b + 6ab^{2}$$

$$8a^{3} – b^{3}- 12a^{2}b + 6ab^{2}\\ = (2a)^{3} – b^{3} – 3(2a)^{2}b + 3(2a)(b)^{2}\\ = (2a – b)^{3}\\ = (2a – b)(2a – b)(2a – b)$$
(iii)27 – 125a3 – 135a + 225a2

$$27 – 125a^{3} – 135a + 225a^{2}\\ = 3^{3} – (5a)^{3} – 3(3)^{2}(5a) + 3(3)(5a)^{2}\\ = (3 – 5a)^{3}\\ = (3 – 5a)(3 – 5a)(3 – 5a)$$

(iv) $$64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}$$

$$64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}\\ = (4a)^{3} – (3b)^{3} – 3(4a)^{2}(3b) + 3(4a)(3b)^{2}\\ = (4a – 3b)^{3}\\ = (4a – 3b)(4a – 3b)(4a – 3b)$$
(v) $$27p^{3} – \frac{1}{216}- \frac{9}{2} p^{2} + \frac{1}{4} p$$

$$27p^{3} – \frac{1}{216}- \frac{9}{2} p^{2} + \frac{1}{4} p$$

$$= (3p)^{3} – (\frac{1}{6})^{3} – 3(3p)^{2}(\frac{1}{6}) + 3(3p)(\frac{1}{6})^{2}\\ = (3p – \frac{1}{6})^{3}\\ = (3p – \frac{1}{6})(3p – \frac{1}{6})(3p – \frac{1}{6})$$

Q.9. Verify:

(i) $$x^{3} + y^{3}\\ = (x + y) (x^{2} – xy + y^{2})$$

(ii) $$x^{3} – y^{3}\\ = (x – y) (x^{2} + xy + y^{2})$$

(i) $$x^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2})\\ We \: know\: that, \\ (x + y)^{3} = x^{3} + y^{3} + 3xy(x + y)\\ \Rightarrow x^{3} +y^{3} = (x + y)^{3} – 3xy(x + y)\\ \Rightarrow x^{3} +y^{3} = (x + y)[(x + y)^{2} – 3xy] {Taking\: (x+y) \: common}\\ \Rightarrow x^{3} +y^{3}= (x + y)[(x^{2} + y^{2} + 2xy) – 3xy] \\ \Rightarrow x^{3} +y^{3} = (x + y)(x^{2} + y^{2} – xy)$$
(ii) $$x^{3} – y^{3}\\ = (x – y) (x^{2} + xy + y^{2})$$
$$x^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2})\\ We \: know\: that, \\ (x – y)^{3} = x^{3} – y^{3} – 3xy(x – y)\\ \Rightarrow x^{3} -y^{3} = (x – y)^{3} + 3xy(x – y)\\ \Rightarrow x^{3} -y^{3} = (x – y)[(x – y)^{2} + 3xy] {Taking\: (x+y) \: common}\\ \Rightarrow x^{3} -y^{3}= (x – y)[(x^{2} + y^{2} – 2xy) + 3xy] \\ \Rightarrow x^{3} +y^{3} = (x – y)(x^{2} + y^{2} + xy)$$

Q.10. Factorize each of the following:

(i) $$27y^{3} + 125z^{3}$$

(ii) $$64m^{3} – 343n^{3}$$

(i) $$27y^{3} + 125z^{3}$$

$$27y^{3} + 125z^{3}\\ = (3y)^{3} + (5z)^{3}\\ = (3y + 5z) [{(3y)^{2} – (3y)(5z) + (5z)^{2}}]\\ = (3y + 5z) (9y^{2} – 15yz + 25z)^{2}$$
(ii) $$64m^{3} – 343n^{3}$$

$$64m^{3} – 343n^{3} \\ = (4m)^{3} – (7n)^{3}\\ = (4m + 7n) [{(4m)^{2} + (4m)(7n) + (7n)^{2}}]\\ = (4m + 7n) (16m^{2} + 28mn + 49n)^{2}$$

Q.11. Factorise : $$27x^{3} + y^{3} + z^{3} – 9xyz$$

$$27x^{3} + y^{3} + z^{3} – 9xyz \\ = (3x)^{3} + y^{3} + z^{3} – 3 (3x)(y)(z)\\ = (3x + y + z) {(3x)^{2} + y^{2} + z^{2} – 3xy – yz – 3xz}\\ = (3x + y + z) (9x^{2} + y^{2} + z^{2} – 3xy – yz – 3xz)$$
Q.12. Verify that:

$$x^{3} + y^{3} + z^{3} – 3xyz = \frac{1}{2}(x + y + z) [(x – y)^{2} + (y – z)^{2} + (z – x)^{2}]$$

We know that,
$$x^{3} + y^{3} + z^{3} -3xyz = (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz – xz)\\ \Rightarrow x^{3} + y^{3} + z^{3} – 3xyz = \frac{1}{2}\times (x + y + z) [2(x^{2} + y^{2} + z^{2} – xy – yz – xz)]\\ = \frac{1}{2}(x + y + z) (2x^{2} + 2y^{2} + 2z^{2} – 2xy – 2yz – 2xz) \\ = \frac{1}{2}(x + y + z) [(x^{2} + y^{2} -2xy) + (y^{2} + z^{2} – 2yz) + (x^{2} + z^{2} – 2xz)] \\ = \frac{1}{2}(x + y + z) [(x – y)^{2} + (y – z)^{2} + (z – x)^{2}]$$

Q.13. If x + y + z = 0, show that $$x^{3} + y^{3} + z^{3} = 3xyz.$$

We know that,
$$x^{3} + y^{3} + z^{3} = 3xyz$$= (x + y + z)(x2 + y2 + z2 – xy – yz – xz)
Now put (x + y + z) = 0,
$$x^{3} + y^{3} + z^{3} = 3xyz= (0)(x^{2} + y^{2}+ z^{2} – xy – yz – xz) \\ \Rightarrow x^{3} + y^{3} + z^{3} – 3xyz = 0$$

Q.14. Without actually calculating the cubes, find the value of each of the following:
(i) $$(-12)^{3} + (7)^{3} + (5)^{3}$$

(ii) $$(28)^{3} +(-15)^{3}+(-13)^{3}$$

(i) $$(-12)^{3} + (7)^{3} + (5)^{3}$$
$$(-12)^{3} + (7)^{3} + (5)^{3}= 0+3(-12)(7)(5)………………(Since. (-12)+(7)+(5)=0)Using\; Identity\\ = -1260$$
(ii) $$(28)^{3} +(-15)^{3}+(-13)^{3}$$
x + y + z = 28 – 15 – 13 = 0

$$(28)^{3}+ (-15)^{3} + (-13)^{3}=0+ 3(28)(-15)(-13) = 16380$$

Q.15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area : $$25a^{2} – 35a + 12$$

(ii) Area : $$35 y^{2} + 13y – 12$$

(i) Area : $$25a^{2} – 35a + 12$$
$$25a^{2} – 35a + 12\\ = 25a^{2} – 15a -20a + 12\\ = 5a(5a – 3) – 4(5a – 3)\\ = (5a – 4)(5a – 3)$$

Possible expression for length = 5a – 4
Possible expression for breadth = 5a – 3

(ii) Area : $$35 y^{2} + 13y – 12$$

$$35 y^{2} + 13y – 12\\ = 35y^{2} – 15y + 28y – 12\\ = 5y(7y – 3) + 4(7y – 3)\\ = (5y + 4)(7y – 3)$$

Possible expression for length = (5y + 4)
Possible expression for breadth = (7y – 3)

Q.16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

(i) Volume : $$3x^{2} – 12x$$
(ii) Volume : $$12ky^{2} + 8ky – 20k$$

(i) Volume : $$3x^{2} – 12x$$
$$3x^{2} – 12x$$
= 3x(x – 4)
Possible expression for length = 3
Possible expression for breadth = x
Possible expression for height = (x – 4)

(ii) Volume : $$12ky^{2} + 8ky – 20k$$

$$12ky^{2} + 8ky – 20k\\ = 4k(3y^{2} + 2y – 5)\\ = 4k(3y^{2} +5y – 3y – 5)\\ = 4k[y(3y +5) – 1(3y + 5)]\\ = 4k (3y +5) (y – 1)$$

Possible expression for length = 4k
Possible expression for breadth = (3y +5)
Possible expression for height = (y -1)