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

(i) 4x² - 3x + 7

(ii) y² + $\sqrt{2}$

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

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

(v) x¹⁰ + y³ + t⁵⁰

Answer

(i) 4x² - 3x + 7
It's a polynomial in one variable i.e., x

(ii) y² + $\sqrt{2}$
It's a polynomial in one variable i.e., y

(iii) $3\sqrt{t}$ + $t\sqrt{2}$ = 3t^1/2 + t $\sqrt{2}$
It is not a polynomial as degree of t is not a whole number.

(iv) y + $\dfrac{2}{y}$ = y + 2y^-1
It is in not a polynomial as degree cannot be negative (-ve) in polynomial.

(v) x¹⁰ + y³ + t⁵⁰
It is polynomial but not in one variable as variables are x, y and t here. Hence, it is a polynomial in three variables.

Write the coefficients of x² in each of the following:

(i) 2 + x² + x

(ii) 2 - x² + x³

(iii) $\dfrac{π}{2}$ x² + x

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

Answer

(i) 2 + x² + x

Coefficient of x² is 1.

(ii) 2 - x² + x³

Coefficient of x² is (-1).

(iii) $\dfrac{π}{2}$ x² + x

Coefficient of x² is = $\dfrac{π}{2}$

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

Coefficient of x² is 0.

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

Answer

One example of a binomial of degree 35 is:

3y³⁵ + 5

One example of a monomial of degree 100 is:

2x¹⁰⁰

Write the degree of each of the following polynomials:

(i) 5x³ + 4x² +7x

(ii) 4 - y²

(iii) 5t - $\sqrt{7}$

(iv) 3

Answer

(i) 5x³ + 4x² +7x

Degree of polynomial is 3.

(ii) 4 - y²

Degree of polynomial is 2.

(iii) 5t - $\sqrt{7}$

Degree of polynomial is 1.

(iv) 3

Degree of polynomial is 0. Degree of constant term is always zero.

Classify the following as linear, quadratic and cubic polynomials:

(i) x² + x

(ii) x - x³

(iii) y + y² + 4

(iv) 1 + x

(v) 3t

(vi) r²

(vii) 7x³

Answer

(i) x² + x — Quadratic polynomial [∵ Degree is 2]

(ii) x - x³ — Cubic polynomial [∵ Degree is 3]

(iii) y + y² + 4 — Quadratic polynomial [∵ Degree is 2]

(iv) 1 + x — Linear polynomial [∵ Degree is 1]

(v) 3t — Linear polynomial [∵ Degree is 1]

(vi) r² — Quadratic polynomial [∵ Degree is 2]

(vii) 7x³ — Cubic polynomial [∵ Degree is 3]

Find the value of the polynomial 5x - 4x² + 3 at

(i) x = 0

(ii) x = -1

(iii) x = 2

Answer

p(x) = 5x - 4x² + 3

(i) x = 0

We put x = 0 in the place of x in p(x)

p(0) = 5(0) - 4(0)² +3

= 0 + 0 + 3

p(0) = 3

Hence, at x = 0, 5x - 4x² + 3 = 3

(ii) x = -1

We put x = -1 in the place of x in p(x)

p(-1) = 5(-1) - 4(-1)² + 3

= (-5) (-4) + 3

= (-9) + 3

= -6

Hence, at x = -1, 5x - 4x² + 3 = -6

(iii) x = 2

We put x = 2 in the place of x in p(x)

p(2) = 5(2) - 4(2)² +3

= 10 - 4 x 4 + 3

= 10 - 16 + 3

= 13 - 16

= -3

Hence, at x = 2, 5x - 4x² + 3 = -3

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

(i) p(y) = y² - y + 1

(ii) p(t) = 2 + t + 2t² - t³

(iii) p(x) = x³

(iv) p(x) = (x - 1) (x + 1)

Answer

(i) p(y) = y² - y + 1

Putting y = 0 we get,

p(0) = (0)² - 0 + 1

= 0 - 0 + 1

= 1

∴ p(0) = 1

Putting y = 1 we get,

p(1) = (1)² - 1 + 1

= 1 - 1 + 1

= 2 - 1

= 1

∴ p(1) = 1

Putting y = 2 we get,

p(2) = (2)² - 2 + 1

= 4 - 2 + 1

= 5 - 2

= 3

∴ p(2) = 1

Hence, for p(y) = y² - y + 1, p(0) = 1, p(1) = 1 and p(2) = 1

(ii) p(t) = 2 + t + 2t² - t³

Putting t = 0 we get,

p(0) = 2 + 0 + 2(0²) - 0³

= 2 + 0 + 0 - 0

= 2

∴ p(0) = 2

Putting t = 1 we get,

p(1) = 2 + 1 + 2(1²) - 1³

= 2 + 1 + 2 - 1

= 4

∴ p(1) = 4

Putting t = 2 we get,

p(2) = 2 + 2 + 2(2²) - 2³

= 4 + 2(4) - 8

= 4 + 8 - 8

= 4

∴ p(2) = 4

Hence, for p(t) = 2 + t + 2t² - t³, p(0) = 2, p(1) = 4 and p(2) = 4

(iii) p(x) = x³

Putting x = 0 we get,

p(0) = (0)³

= 0

∴ p(0) = 0

Putting x = 1 we get,

p(1) = (1)³

= 1

∴ p(1) = 1

Putting x = 2 we get,

p(2) = (2)³

= 8

∴ p(2) = 8

Hence, for p(x) = x³, p(0) = 0, p(1) = 1 and p(2) = 8

(iv) p(x) = (x - 1)(x + 1)

Putting x = 0 we get,

p(0) = (0 - 1)(0 + 1)

= (-1) x 1

= (-1)

∴ p(0) = -1

Putting x = 1 we get,

p(1) = (1 - 1)(1 + 1)

= 0 x 2

= 0

∴ p(1) = 0

Putting x = 2 we get

p(2) = (2 - 1) (2 + 1)

= 1 x 3

= 3

∴ p(2) = 3

Hence, for p(x) = (x - 1) (x + 1), p(0) = -1, p(1) = 0 and p(2) = 3

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

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

(ii) p(x) = 5x - π, x = $\dfrac{4}{5}$

(iii) p(x) = x² - 1, x = 1, -1

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

(v) p(x) = x², x = 0

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

(vii) p(x) = 3x² - 1, x = $-\Big(\dfrac{1}{\sqrt{3}}\Big)$, $\Big(\dfrac{2}{\sqrt{3}}\Big)$

(viii) p(x) = 2x + 1, x = $\dfrac{1}{2}$

Answer

(i) p(x) = 3x + 1

Putting x = $-\dfrac{1}{3}$ we get,

$\text{p}\Big(-\dfrac{1}{3}\Big)$ = $3\Big(-\dfrac{1}{3}\Big)$ + 1

= -1 + 1

= 0

Hence, x = $-\dfrac{1}{3}$ is zeroes of this polynomial.

(ii) p(x) = 5x - π

Putting x = $\dfrac{4}{5}$ we get,

$\text{p}\Big(\dfrac{4}{5}\Big) = 5\Big(\dfrac{4}{5}\Big) - \pi$

= 4 - π

Hence, x = $\dfrac{4}{5}$ is not a zeroes of this polynomial.

(iii) p(x) = x² - 1

Putting x = 1 we get,

p(1) = (1)² - 1

= 0

Putting x = -1 we get

p(-1) = (-1)² - 1

= 1 - 1

= 0

Hence, x = 1, -1 is zeroes of this polynomial.

(iv) p(x) = (x + 1)(x - 2)

Putting x = -1 we get,

p(1) = (-1 + 1)(-1 - 2)

= 0 x (-3)

= 0

Putting x = 2 we get,

p(2) = (2 + 1)(2 - 2)

= 3 x 0

= 0

Hence, x = -1, 2 are the zeroes of this polynomial.

(v) p(x) = x²

Putting x = 0 we get,

p(0) = (0)²

= 0

Hence, x = 0 is the zero of this polynomial.

(vi) p(x) = lx + m

Putting x = -$\dfrac{m}{l}$ we get,

p - $\Big(\dfrac{m}{l}\Big) = l x -\Big(\dfrac{m}{l}\Big)$ + m

= -m + m

= 0

Hence, x = $-\dfrac{m}{l}$ is the zero of this polynomial.

(vii) p(x) = 3x² - 1

Putting x = $-\dfrac{1}{\sqrt{3}}$ we get,

p-$\Big(\dfrac{1}{\sqrt{3}}\Big)$ = 3 x -$\Big(\dfrac{1}{\sqrt{3}}\Big)$² - 1

= 3 x $\dfrac{1}{3}$ - 1

= 1 - 1

= 0

Putting x = $\dfrac{2}{\sqrt{3}}$ we get,

p $\Big(\dfrac{2}{\sqrt{3}}\Big)$ = 3 x $\Big(\dfrac{2}{\sqrt{3}}\Big)$² - 1

= 3 x $\dfrac{4}{3}$ - 1

= 4 - 1

= 3

Hence, x = $-\dfrac{1}{\sqrt{3}}$ is zeroes of polynomial but x = $\dfrac{2}{\sqrt{3}}$ is not a zeroes of this polynomial.

(viii) p(x) = 2x + 1

Putting x = $\dfrac{1}{2}$ we get,

p$\Big(\dfrac{1}{2}\Big)$ = 2 x $\dfrac{1}{2}$ + 1

= 1 + 1

= 2

No, x = $\dfrac{1}{2}$ is not the zeroes of this polynomial.

Find the zero of the polynomial in each of the following cases:

(i) p(x) = x + 5

(ii) p(x) = x - 5

(iii) p(x) = 2x + 5

(iv) p(x) = 3x - 2

(v) p(x) = 3x

(vi) p(x) = ax, a ≠ 0

(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.

Answer

(i) p(x) = x + 5

⇒ p(x) = 0

⇒ x + 5 = 0

⇒ x = -5

Hence, -5 is a zero of polynomial x + 5

(ii) p(x) = x - 5

⇒ p(x) = 0

⇒ x - 5 = 0

⇒ x = 5

Hence, 5 is a zero of polynomial x - 5

(iii) p(x) = 2x + 5

⇒ p(x) = 0

⇒ 2x + 5 = 0

⇒ 2x = -5

⇒ x = $-\dfrac{5}{2}$

Hence, $-\dfrac{5}{2}$ is a zero of polynomial 2x + 5

(iv) p(x) = 3x - 2

⇒ p(x) = 0

⇒ 3x - 2 = 0

⇒ 3x = 2

⇒ x = $\dfrac{2}{3}$

Hence, $\dfrac{2}{3}$ is a zero of polynomial 3x - 2

(v) p(x) = 3x

⇒ p(x) = 0

⇒ 3x = 0

⇒ x = $\dfrac{0}{3}$

⇒ x = 0

Hence, 0 is a zero of polynomial 3x

(vi) p(x) = ax

⇒ p(x) = 0

⇒ ax = 0

⇒ x = $\dfrac{0}{a}$

⇒ x = 0

Hence, 0 is a zero of polynomial ax

(vii) p(x) = cx + d

⇒ p(x) = 0

⇒ cx + d = 0

⇒ cx = -d

x = $\dfrac{-d}{c}$

Hence, $-\dfrac{d}{c}$ is a zero of polynomial cx + d

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

(i) x³ + x² + x + 1

(ii) x⁴ + x³ + x² + x + 1

(iii) x⁴ + 3x³ + 3x² + x + 1

(iv) x³ - x² - (2 + $\sqrt{2}$)x + $\sqrt{2}$

Answer

(i) x³ + x² + x + 1

⇒ x + 1 = 0

⇒ x = -1

p(x) = x³ + x² + x + 1

p(-1) = (-1)³ + (-1)² + (-1) + 1

= -1 + 1 -1 + 1

= 0

Remainder is zero (0), so (x + 1) is factor of this polynomial.

(ii) x⁴ + x³ + x² + x + 1

⇒ x + 1 = 0

⇒ x = -1

p(x) = x⁴ + x³ + x² + x + 1

p(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1) + 1

= 1 -1 + 1 -1 + 1

= 1

Remainder is not zero (0), so (x + 1) is not a factor of this polynomial.

(iii) x⁴ + 3x³ + 3x² + x + 1

⇒ x + 1 = 0

⇒ x = -1

p(x) = x⁴ + 3x³ + 3x² + x + 1

p(-1) = (-1)⁴ + 3 x (-1)³ + 3 x (-1²) + (-1) + 1

= 1 -3 + 3 -1 + 1

= 1

Remainder is not zero (0), so (x + 1) is not a factor of this polynomial.

(iv) x³ - x² - (2 + $\sqrt{2}$)x + $\sqrt{2}$

⇒ x + 1 = 0

⇒ x = -1

p(x) = x³ - x² - (2 + $\sqrt{2}$)x + $\sqrt{2}$

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

= -1 - 1 + 2 + $\sqrt{2}$ + $\sqrt{2}$

= -2 + 2 + 2 $\sqrt{2}$

= 2 $\sqrt{2}$

Remainder is not zero (0), so (x + 1) is not a factor of this polynomial.

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

(i) p(x) = 2x³ + x² - 2x - 1, g(x) = x + 1

(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2

(iii) p(x) = x³ - 4x² + x + 6, g(x) = x - 3

Answer

(i) p(x) = 2x³ + x² - 2x - 1

g(x) = x + 1

⇒ x + 1 = 0

⇒ x = -1

Putting x = -1 we get,

p(-1) = 2(-1)³ + (-1)² - 2(-1) - 1

= -2 + 1 + 2 -1

= 0

Remainder is zero (0), so g(x) is factor of p(x).

(ii) p(x) = x³ + 3x² + 3x + 1

g(x) = x + 2

⇒ x + 2 = 0

⇒ x = -2

Putting x = -2 we get,

p(-2) = (-2)³ + 3(-2)² + 3(-2) + 1

= -8 + 3(4) - 6 + 1

= -8 + 12 - 6 + 1

= 13 - 14

= -1

Remainder is not zero (0), so g(x) is not a factor of p(x).

(iii) p(x) = x³ - 4x² + x + 6

g(x) = x - 3

⇒ x - 3 = 0

⇒ x = 3

Putting x = -3 we get,

p(3) = (3)³ - 4(3)² + 3 + 6

= 27 - 4(9) + 9

= 27 - 36 + 9

= 36 - 36

= 0

Remainder is zero (0), so g(x) is a factor of p(x).

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

(i) p(x) = x² + x + k

(ii) p(x) = 2x² + kx + $\sqrt{2}$

(iii) p(x) = kx² - $\sqrt{2}$x + 1

(iv) p(x) = kx² - 3x + k

Answer

(i) p(x) = x² + x + k

⇒ x - 1 = 0

⇒ x = 1

Putting x = 1 we get,

p(1) = (1)² + 1 + k

= 1 + 1 + k

⇒ 2 + k = 0

⇒ k = -2

So the value of k = -2

(ii) p(x) = 2x² + kx + $\sqrt{2}$

⇒ x - 1 = 0

⇒ x = 1

Putting x = 1 we get,

p(1) = 2(1)² + k(1) + $\sqrt{2}$

⇒ 2 + k + $\sqrt{2}$

⇒ k = -2 -$\sqrt{2}$

⇒ k = -(2 + $\sqrt{2}$)

So the value of k = -(2 + $\sqrt{2}$)

(iii) p(x) = kx² - $\sqrt{2}$x + 1

⇒ x - 1 = 0

⇒ x = 1

Putting x = 1 we get,

p(1) = k(1)² - $\sqrt{2}$(1) + 1

⇒ k - $\sqrt{2}$ + 1 = 0

⇒ k = $\sqrt{2}$ - 1

So the value of k = $\sqrt{2}$ - 1

(iv) p(x) = kx² - 3x + k

⇒ x - 1 = 0

⇒ x = 1

Putting x = 1 we get,

p(1) = k(1)² -3(1) + k

= k - 3 + k

⇒ 2k - 3 = 0

⇒ 2k = 3

⇒ k = $\dfrac{3}{2}$

So the value of k = $\dfrac{3}{2}$

Factorise :

(i) 12x² - 7x + 1

(ii) 2x² + 7x + 3

(iii) 6x² + 5x - 6

(iv) 3x² - x - 4

Answer

(i) 12x² - 7x + 1

= 12x² - (4 + 3)x + 1

= 12x² -4x -3x + 1

= 4x(3x - 1) -1(3x - 1)

= (3x - 1)(4x - 1)

Hence, 12x² - 7x + 1 = (3x - 1)(4x - 1)

(ii) 2x² + 7x + 3

= 2x² + (6 + 1)x + 3

= 2x² + 6x + x + 3

= 2x(x + 3) + 1(x + 3)

= (2x + 1)(x + 3)

Hence, 2x² + 7x + 3 = (2x + 1)(x + 3)

(iii) 6x² + 5x - 6

= 6x² + (9 - 4)x - 6

= 6x² +9x - 4x - 6

= 3x(2x + 3) -2(2x + 3)

= (2x + 3)(3x - 2)

Hence, 6x² + 5x - 6 = (2x + 3)(3x - 2)

(iv) 3x² - x - 4

= 3x² - (4 - 3)x - 4

= 3x² - 4x + 3x - 4

= x(3x - 4) + 1(3x - 4)

= (3x - 4)(x + 1)

Hence, 3x² - x - 4 = (3x - 4)(x + 1)

Factorise :

(i) x³ - 2x² - x + 2

(ii) x³ - 3x² - 9x - 5

(iii) x³ + 13x² + 32x + 20

(iv) 2y³ + y² - 2y - 1

Answer

(i) x³ - 2x² - x + 2

Let (x - a) is a factor, a = 1, -1, 2, -2.......

Putting a = 1

x - 1 = 0

x = 1

p(x) = x³ - 2x² - x + 2

p(1) = (1)³ - 2 x (1)² - 1 + 2

= 1 - 2 - 1 + 2

= 0

Hence, (x - 1) is the factor of x³ - 2x² - x + 2

On dividing, x³ - 2x² - x + 2 by (x - 1)

$\begin{array}{l} \phantom{x - 1)}{\quad x^2 -x - 2} \\ x - 1\overline{\smash{\big)}\quad x^3 - 2x^2 - x + 2} \\ \phantom{x - 1)}\phantom{2}\underline{\underset{-}{+}x^3 \underset{+}{-}x^2} \\ \phantom{{x - 1}x^3-2}-x^2 - x + 2 \\ \phantom{{x - 1)}x^3-2}\underline{\underset{+}{-}x^2 \underset{-}{+} x} \\ \phantom{{x - 1)}{x^3-2x^{2}(3)}}-2x + 2 \\ \phantom{{x - 1)}{x^3-2x^{2}(31)}}\underline{\underset{+}{-}2x \underset{-}{+} 2} \\ \phantom{{x - 1)}{x^3-2x^{2}(31)}{-2x}}\times \end{array}$

we get quotient = x² - x - 2

Factorising x² - x - 2,

= x² - x - 2

= x² -(2 - 1)x - 2

= x² - 2x + x -2

= x( x - 2) + 1(x + 2)

= (x + 1)(x - 2)

∴ x² - x - 2 = (x + 1)(x - 2)

∴ p(x) = x³ - 2x² - x + 2 = (x - 1) (x + 1) (x - 2)

So, factors are (x - 1) (x + 1) (x - 2)

(ii) x³ - 3x² - 9x - 5

Let (x - a) is a factor, a = 1, -1, 2, -2.......

Putting a = -1

x + 1 = 0

x = -1

p(x) = x³ - 3x² - 9x - 5

p(-1) = (-1)³ - 3 x (-1)² - 9 x (-1) - 5

= -1 - 3 + 9 - 5

= -9 + 9

= 0

Hence, (x + 1) is a factor of x³ - 3x² - 9x - 5

On dividing, x³ - 3x² - 9x - 5 by (x + 1)

$\begin{array}{l} \phantom{x + 1)}{\quad x^2 -4x - 5} \\ x + 1\overline{\smash{\big)}\quad x^3 - 3x^2 - 9x - 5} \\ \phantom{x - 1)}\phantom{2}\underline{\underset{-}{+}x^3 \underset{-}{+}x^2} \\ \phantom{{x + 1}x^3-2}-4x^2 - 9x \\ \phantom{{x + 1)}x^3-2}\underline{\underset{+}{-}4x^2 \underset{+}{-} 4x} \\ \phantom{{x + 1)}{x^3-2x^{2}(3)}}-5x - 5 \\ \phantom{{x + 1)}{x^3-2x^{2}(31)}}\underline{\underset{+}{-}5x \underset{+}{-} 5} \\ \phantom{{x + 1)}{x^3-2x^{2}(31)}{-2}}\times \end{array}$

We get quotient = x² - 4x - 5

Factorising x² - 4x - 5

x² -(5 - 1)x - 5

x² -5x + x - 5

x(x - 5) + 1(x - 5)

(x + 1) (x - 5)

∴ x² - 4x - 5 = (x + 1) (x - 5)

∴ p(x) = x³ - 3x² -9x - 5 = (x + 1) (x + 1) (x - 5)

So, factors are (x + 1) (x + 1) (x - 5)

(iii) x³ + 13x² + 32x + 20

Let (x - a) is a factor, a = 1, -1, 2, -2.......

Putting a = -1

x + 1 = 0

x = -1

p(x) = x³ + 13x² + 32x + 20

p(-1) = (-1)³ + 13 x (-1)² + 32 x (-1) + 20

= -1 + 13 - 32 + 20

= -33 + 33

= 0

Hence, (x + 1) is a factor of x³ + 13x² + 32x + 20

On dividing, x³ + 13x² + 32x + 20 by (x + 1)

$\begin{array}{l} \phantom{x + 1)}{\quad x^2 +12x + 20} \\ x + 1\overline{\smash{\big)}\quad x^3 + 13x^2 + 32x + 20} \\ \phantom{x - 1)}\phantom{2}\underline{\underset{-}{+}x^3 \underset{-}{+}x^2} \\ \phantom{{x + 1}x^3-2}12x^2 + 32x + 20 \\ \phantom{{x + 1)}x^31}\underline{\underset{-}{+}12x^2 \underset{-}{+} 12x} \\ \phantom{{x + 1)}{x^3-2x^{2}(3)}}20x + 20 \\ \phantom{{x + 1)}{x^3-2x^{2}3}}\underline{\underset{-}{+}20x \underset{-}{+} 20} \\ \phantom{{x + 1)}{x^3-2x^{2}(3)}{-2}}\times \end{array}$

We get quotient = x² + 12x + 20

Factorising x² + 12x + 20

x² + 10x + 2x + 20

x(x + 10) + 2(x + 10)

(x + 10)(x + 2)

∴ x² + 12x + 20 = (x + 10)(x + 2)

∴ p(x) = x³ + 13x² + 32x + 20 = (x + 1) (x + 10) (x + 2)

So, factors are (x + 1) (x + 10) (x + 2)

(iv) 2y³ + y² - 2y - 1

Let (y - a) is a factor, a = 1, -1, 2, -2.......

Putting a = 1

y - 1 = 0

y = 1

p(y) = 2y³ + y² - 2y - 1

p(1) = 2 x (1)³ + (1)² - 2 x 1 - 1

= 2 + 1 - 2 - 1

= 0

Hence, (y - 1) is a factor of 2y³ + y² - 2y - 1

On dividing, 2y³ + y² - 2y - 1 by (y - 1)

$\begin{array}{l} \phantom{y - 1)}{\quad 2y^2 + 3y + 1} \\ y - 1\overline{\smash{\big)}\quad 2y^3 + y^2 - 2y - 1} \\ \phantom{y - 1)}\underline{\underset{-}{+}2y^3 \underset{+}{-}2y^2} \\ \phantom{{y - 1}x^3-233}3y^2 - 2y \\ \phantom{{y - 1)}x^3-2}\underline{\underset{-}{+}3y^2 \underset{+}{-} 3y} \\ \phantom{{y - 1)}{x^3-2x^{2}(3333)}}y - 1 \\ \phantom{{y - 1)}{x^3-2x^{2}3333}}\underline{\underset{-}{+}y \underset{+}{-} 1} \\ \phantom{{y - 1)}{x^3-2x^{2}(333)}{-2}}\times \end{array}$

We get quotient = 2y² + 3y + 1

Factorising 2y² + 3y + 1

2y² + 2y + y +1

2y(y + 1) + 1(y + 1)

(y + 1)(2y + 1)

∴ 2y² + 3y + 1 = (y + 1)(2y + 1)

∴ p(y) = 2y³ + y² - 2y - 1 = (y - 1)(y + 1)(2y + 1)

So, factors are (y - 1)(y + 1)(2y + 1)

Use suitable identities to find the following products:

(i) (x + 4)(x + 10)

(ii) (x + 8)(x - 10)

(iii) (3x + 4)(3x - 5)

(iv) (y² + $\dfrac{3}{2}$) (y² - $\dfrac{3}{2}$)

(v) (3 - 2x)(3 + 2x)

Answer

(i) (x + 4)(x + 10)

[∵ (x + a)(x + b) = x² + (a + b)x + ab]

Putting a = 4, b = 10

= x² + (4 + 10)x + 4 x 10

= x² + 14x + 40

Hence, (x + 4)(x + 10) = x² + 14x + 40

(ii) (x + 8)(x - 10)

[∵ (x + a)(x + b) = x² + (a + b)x + ab]

Putting a = 8 , b = -10

= x² + [8 + (-10)]x + 8 x (-10)

= x² - 2x - 80

Hence, (x + 8)(x - 10) = x² - 2x - 80

(iii) (3x + 4)(3x - 5)

[∵ (x + a)(x + b) = x² + (a + b)x + ab]

Putting a = 4, b = -5, x = 3x

= (3x)² + [4 + (-5)](3x) + (4)(-5)

= 9x² + (-1)(3x) - 20

= 9x² - 3x - 20

Hence, (3x + 4) (3x - 5) = 9x² - 3x - 20

(iv) (y² + $\dfrac{3}{2}$)(y² - $\dfrac{3}{2}$)

[∵ (a + b)(a - b) = a² - b²]

Putting a = y², b = $\dfrac{3}{2}$

= (y²)² - ($\dfrac{3}{2}$)²

= y⁴ - $\dfrac{9}{4}$

Hence, (y² + $\dfrac{3}{2}$) (y² - $\dfrac{3}{2}$) = y⁴ - $\dfrac{9}{4}$

(v) (3 - 2x)(3 + 2x)

[∵ (a - b)(a + b) = a² - b²]

Putting a = 3, b = 2x

= (3)² - (2x)²

= 9 - 4x²

Hence, (3 - 2x)(3 + 2x) = 9 - 4x²

Evaluate the following products without multiplying directly:

(i) 103 × 107

(ii) 95 × 96

(iii) 104 × 96

Answer

(i) 103 × 107

We can write it as,

= (100 + 3)(100 + 7)

[∵ (x + a)(x + b) = x² + (a + b)x + ab]

Putting x = 100, a = 3, b = 7

= (100)² + (3 + 7)(100) + (3)(7)

= 10000 + (10)(100) + 21

= 10000 + 1000 + 21

= 11021

Hence, 103 × 107 = 11021

(ii) 95 × 96

We can write it as,

[100 + (-5)][100 + (-4)]

[∵ (x + a)(x + b) = x² + (a + b)x + ab]

Putting x = 100, a = -5 , b = -4

= (100)² - [(-5) + (-4)](100) + (-5) x (-4)

= 10000 - (9)(100) + 20

= 10000 - 900 + 20

= 10020 - 900

= 9120

Hence, 95 × 96 = 9120

(iii) 104 × 96

We can write it as,

(100 + 4)(100 - 4)

[∵ (a + b)(a - b) = a² - b²]

Putting a = 100, b = 4

= (100)² - (4)²

= 10000 - 16

= 9984

Hence, 104 × 96 = 9984

Factorise the following using appropriate identities:

(i) 9x² + 6xy + y²

(ii) 4y² - 4y + 1

(iii) $x^2 - \Big(\dfrac{y^2}{100}\Big)$

Answer

(i) 9x² + 6xy + y²

[∵ a² + 2ab + b² = (a + b)²]

= (3x)² + 2(3x)(y) + (y)²

= (3x + y)²

Hence, 9x² + 6xy + y² = (3x + y)(3x + y)

(ii) 4y² - 4y + 1

[∵ a² - 2ab + b² = (a - b)²]

= (2y)² - 2(2y)(1) + (1)²

= (2y - 1)²

Hence, 4y² - 4y + 1 = (2y - 1)(2y - 1)

(iii) $x^2 - \Big(\dfrac{y^2}{100}\Big)$

[∵ a² - b² = (a - b)(a + b)]

$x^2 - \Big(\dfrac{y^2}{100}\Big) \\[1em] = x^2 - \Big(\dfrac{y}{10}\Big)^2 \\[1em] = \Big(x - \dfrac{y}{10}\Big) \Big(x + \dfrac{y}{10}\Big)$

Hence, $x^2 - \Big(\dfrac{y^2}{100}\Big)$ = $\Big(x - \dfrac{y}{10}\Big) \Big(x + \dfrac{y}{10}\Big)$

Expand each of the following, using suitable identities:

(i) (x + 2y + 4z)²

(ii) (2x - y + z)²

(iii) (-2x + 3y + 2z)²

(iv) (3a - 7b - c)²

(v) (-2x + 5y - 3z)²

(vi) $\Big[\dfrac{1}{4}a - \dfrac{1}{2}b + 1\Big]^2$

Answer

(i) (x + 2y + 4z)²

[∵ (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx]

Putting x = x, y = 2y, z = 4z

= (x)² + (2y)² + (4z)² + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)

= x² + 4y² + 16z² + 4xy + 16yz + 8zx

(ii) (2x - y + z)²

[∵ (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx]

Putting x = 2x, y = (-y), z = z

= (2x)² + (-y)² + (z)² + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)

= 4x² + y² + z² - 4xy - 2yz + 4zx

(iii) (-2x + 3y + 2z)²

[∵ (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx]

Putting x = -2x, y = 3y, z = 2z

= (-2x)² + (3y)² + (2z)² + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)

= 4x² + 9y² + 4z² - 12xy + 12yz - 8zx

(iv) (3a - 7b - c)²

[∵ (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx]

Putting x = 3a, y = (-7b), z = (-c)

= (3a)² + (-7b)² + (-c)² + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)

= 9a² + 49b² + c² - 42ab + 14bc - 6ac

(v) (-2x + 5y - 3z)²

[∵ (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx]

Putting x = -2x, y = 5y, z = (-3z)

= (-2x)² + (5y)² + (-3z)² + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)

= 4x² + 25y² + 9z² - 20xy - 30yz + 12zx

(vi) $\Big[\dfrac{1}{4}a - \dfrac{1}{2}b + 1\Big]^2$

[∵ (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx]

Putting x = $\dfrac{a}{4}, y = -\dfrac{b}{2}$, z = 1

$= \Big(\dfrac{a}{4}\Big)^2 + \Big(-\dfrac{b}{2}\Big)^2 + (1)^2 + 2\Big(\dfrac{a}{4}\Big) \Big(-\dfrac{b}{2}\Big) + 2\Big(-\dfrac{b}{2}\Big)(1) + 2(1) \Big(\dfrac{a}{4}\Big) \\[1em] = \dfrac{a^2}{16} + \dfrac{b^2}{4} + 1 - \dfrac{ab}{4} -b + \dfrac{a}{2}$

Factorise:

(i) 4x² + 9y² + 16z² + 12xy - 24yz - 16xz

(ii) 2x² + y² + 8z² - $2\sqrt{2}$xy + $4\sqrt{2}$yz - 8xz

Answer

(i) 4x² + 9y² + 16z² + 12xy - 24yz - 16xz

= (2x)² + (3y)² + (-4z)² + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)

= (2x + 3y - 4z)² [∵ (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx]

Hence, 4x² + 9y² + 16z² + 12xy - 24yz - 16xz = (2x + 3y - 4z)(2x + 3y - 4z)

(ii) 2x² + y² + 8z² - $2\sqrt{2}$xy + $4\sqrt{2}$yz - 8xz

[∵ (a + b + c)² = (a)² + (b)² + (c)² + 2ab + 2bc + 2ca]

$= (-\sqrt{2}x)^2 + (y)^2 + (2\sqrt{2}z)^2 + 2(-\sqrt{2}x)(y) + 2(y)(2\sqrt{2}z) + 2(2\sqrt{2}z)(-\sqrt{2}x) \\[1em] = (-\sqrt{2}x + y + 2\sqrt{2}z)^2$

Hence, 2x² + y² + 8z² -$2\sqrt{2}$xy + $4\sqrt{2}$yz - 8zx = (-$\sqrt{2}$x + y + $2\sqrt{2}$z)(-$\sqrt{2}$x + y + $2\sqrt{2}$z)

Write the following cubes in expanded form:

(i) (2x + 1)³

(ii) (2a - 3b)³

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

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

Answer

(i) (2x + 1)³

We know that:

(a + b)³= (a)³ + (b)³ + 3a²b + 3ab²

Putting a = 2x, b = 1

= (2x)³ + (1)³ + 3(2x)²(1) + 3(2x)(1)²

= 8x³ + 12x² + 6x + 1

Hence, (2x + 1)³ = 8x³ + 12x² + 6x + 1

(ii) (2a - 3b)³

We know that:

(a - b)³= (a)³ - (b)³ - 3a²b + 3ab²

Putting a = 2a, b = 3b

= (2a)³ - (3b)³ - 3(2a)²(3b) + 3(2a)(3b)²

= 8a³ - 27b³ - 3(4a²)(3b) + 3(2a)(9b²)

= 8a³ - 27b³ - 36a²b + 54ab²

Hence, (2a - 3b)³ = 8a³ - 27b³ - 36a²b + 54ab²

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

We know that:

(a + b)³ = (a)³ + (b)³ + 3ab(a + b)

Putting a = $\dfrac{3}{2}x$, b = 1

$= \Big(\dfrac{3}{2}\Big)^3 + (1)^3 + 3 \Big(\dfrac{3x}{2}\Big)(1) \Big(\dfrac{3x}{2} + 1\Big) \\[1em] = \dfrac{27x^3}{8} + 1 + \Big(\dfrac{9x}{2}\Big) \Big(\dfrac{3x}{2} + 1\Big) \\[1em] = \dfrac{27x^3}{8} + 1 + \Big(\dfrac{9x}{2}\Big) \Big(\dfrac{3x}{2}\Big) + \dfrac{9x}{2}(1)\\[1em] = \dfrac{27x^3}{8} + \dfrac{27x^2}{4} + \dfrac{9x}{2} + 1$

Hence, $\Big[\dfrac{3x}{2} + 1\Big]^3 = \dfrac{27x^3}{8} + \dfrac{27x^2}{4} + \dfrac{9x}{2} + 1$

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

We know that:

(a - b)³ = (a)³ - (b)³ - 3ab(a - b)

Putting a = x, b = $-\dfrac{2}{3}y$

$= (x)^3 - \Big[-\dfrac{2}{3}y\Big]^3 - 3(x)\Big[-\dfrac{2}{3}y\Big] \Big[x - \dfrac{2}{3}y\Big]\\[1em] = x^3 - \dfrac{8}{27}y^3 - 2xy\Big[x - \dfrac{2}{3}y\Big]\\[1em] = x^3 - \dfrac{8}{27}y^3 - 2x^2y + \dfrac{4}{3}xy^2$

Hence, $\Big[x - \dfrac{2}{3}y\Big]^3 = x^3 - \dfrac{8}{27}y^3 - 2x^2y + \dfrac{4}{3}xy^2$

Evaluate the following using suitable identities:

(i) (99)³

(ii) (102)³

(iii) (998)³

Answer

(i) (99)³

We can write it as,

= (100 - 1)³

We know that,

(a - b)³ = (a)³ - (b)³ - 3ab(a - b)

Putting a = 100 , b = 1

= (100)³ - (1)³ - 3 x 100 x 1(100 - 1)

= 1000000 - 1 - 300 x 99

= 1000000 - 1 - 29700

= 970299

Hence, (99)³ = 970299

(ii) (102)³

We can write it as,

= (100 + 2)³

We know that,

(a + b)³ = (a)³ + (b)³ + 3ab(a + b)

Putting a = 100 , b = 2

= (100)³ + (2)³ + 3 x 100 x 2(100 + 2)

= 1000000 + 8 + 600 x 102

= 1000000 + 8 + 61200

= 1061208

Hence, (102)³ = 1061208

(iii) (998)³

We can write it as,

= (1000 - 2)³

We know that,

(a - b)³ = (a)³ - (b)³ - 3ab(a - b)

Putting a = 1000 , b = 2

= (1000)³ - (2)³ - 3 x 1000 x 2(1000 - 2)

= 1000000000 - 8 - 6000 x 998

= 1000000000 - 8 - 5988000

= 994011992

Hence, (998)³ = 994011992

Factorise each of the following:

(i) 8a³ + b³ + 12a²b + 6ab²

(ii) 8a³ - b³ - 12a²b + 6ab²

(iii) 27 - 125a³ - 135a + 225a²

(iv) 64a³ - 27b³ - 144a²b + 108ab²

(v) 27p³ - $\dfrac{1}{216}$ - $\dfrac{9}{2}$p² + $\dfrac{1}{4}$p

Answer

(i) 8a³ + b³ + 12a²b + 6ab²

[∵ a³ + b³ + 3a²b + 3ab² = (a + b)³]

= (2a)³ + (b)³ + 3(2a)²(b) + 3(2a)(b)²

= (2a + b)³

Hence, 8a³ + b³ + 12a²b + 6ab² = (2a + b)(2a + b)(2a + b)

(ii) 8a³ - b³ - 12a²b + 6ab²

[∵ a³ - b³ - 3a²b + 3ab² = (a - b)³]

= (2a)³ - (b)³ - 3(2a)²(b) + 3(2a)(b)²

= (2a - b)³

Hence, 8a³ - b³ - 12a²b + 6ab² = (2a - b)(2a - b)(2a - b)

(iii) 27 - 125a³ - 135a + 225a²

[∵ a³ - b³ - 3a²b + 3ab² = (a - b)³]

= (3)³ - (5a)³ - 3(3)²(5a) + 3(3)(5a)²

= (3 - 5a)³

Hence, 27 - 125a³ - 135a + 225a² = (3 - 5a)(3 - 5a)(3 - 5a)

(iv) 64a³ - 27b³ - 144a²b + 108ab²

[∵ a³ - b³ - 3a²b + 3ab² = (a - b)³]

= (4a)³ - (3b)³ - 3(4a)²(3b) + 3(4a)(3b)²

= (4a - 3b)³

Hence, 64a³ - 27b³ - 144a²b + 108ab² = (4a - 3b)(4a - 3b)(4a - 3b)

(v) 27p³ - $\dfrac{1}{216}$ - $\dfrac{9}{2}$p² + $\dfrac{1}{4}$p

[∵ a³ - b³ - 3a²b + 3ab² = (a - b)³]

= (3p)³ - $\Big(\dfrac{1}{6}\Big)$³ - 3(3p)²$\Big(\dfrac{1}{6}\Big)$ + 3(3p)$\Big(\dfrac{1}{6}\Big)$²

= $\Big(3p - \dfrac{1}{6}\Big)^3$

Hence, 27p³ - $\dfrac{1}{216}$ - $\dfrac{9}{2}$p² + $\dfrac{1}{4}$p = $\Big(3p - \dfrac{1}{6}\Big)$\Big(3p - \dfrac{1}{6}\Big)$\Big(3p - \dfrac{1}{6}\Big)$

Verify :

(i) x³ + y³ = (x + y) (x² - xy + y²)

(ii) x³ - y³ = (x - y) (x² + xy + y²)

Answer

(i) x³ + y³ = (x + y) (x² - xy + y²)

R.H.S

= (x + y) (x² - xy + y²)

= (x)(x² - xy + y²) + (y)(x² - xy + y²)

= x³ - x²y + xy² + yx² - xy² + y³

= x³ + y³

Hence, L.H.S = R.H.S

(ii) x³ - y³ = (x - y) (x² + xy + y²)

R.H.S

= (x - y) (x² + xy + y²)

= (x)(x² + xy + y²) - (y)(x² + xy + y²)

= x³ + x²y + xy² - yx² - xy² - y³

= x³ - y³

Hence, L.H.S = R.H.S

Factorise each of the following:

(i) 27y³ + 125z³

(ii) 64m³ - 343n³

Answer

(i) 27y³ + 125z³

[∵ x³ + y³ = (x + y)(x² - xy + y²)]

⇒ 27y³ + 125z³ = (3y)³ + (5z)³

Putting x = 3y, y = 5z

R.H.S = (x + y)(x² - xy + y²)

= (3y + 5z)[(3y)² - (3y)(5z) + (5z)²]

= (3y + 5z)(9y² - 15yz + 25z²)

Hence, 27y³ + 125z³ = (3y + 5z)(9y² - 15yz + 25z²)

(ii) 64m³ - 343n³

[∵ a³ - b³ = (a - b)(a² + ab + b²)]

⇒ 64m³ - 343n³ = (4m)³ - (7n)³

Putting a = 4m, b = 7n

R.H.S = (a - b)(a² + ab + b²)

= (4m - 7n)[(4m)² + (4m)(7n) + (7n)²]

= (4m - 7n)(16m² + 28mn + 49n²)

Hence, 64m³ - 343n³ = (4m - 7n)(16m² + 28mn + 49n²)

Factorise : 27x³ + y³ + z³ - 9xyz

Answer

27x³ + y³ + z³ - 9xyz

[∵ x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)]

= (3x)³ + y³ + z³ - 3(3x)(y)(z)

= (3x + y + z)[(3x)² + y² + z² - (3x)(y) - (y)(z) - (z)(3x)]

= (3x + y + z)(9x² + y² + z² - 3xy - yz - 3zx)

Hence, 27x³ + y³ + z³ - 9xyz = (3x + y + z)(9x² + y² + z² - 3xy - yz - 3zx)

Verify that x³ + y³ +z³ - 3xyz = $\dfrac{1}{2}$(x + y + z)[(x - y)² + (y - z)² + (z - x)²]

Answer

R.H.S

= $\dfrac{1}{2}$(x + y + z)[(x - y)² + (y - z)² + (z - x)²]

= $\dfrac{1}{2}$(x + y + z)[(x² + y² - 2xy) + (y² + z² - 2yz) + (z² + x² - 2zx)] [∵ (a - b)² = (a)² + (b)² - 2ab]

= $\dfrac{1}{2}$(x + y + z)[2x² + 2y² + 2z² - 2xy - 2yz - 2zx]

= $\dfrac{1}{2}$(x + y + z) 2[x² + y² + z² - xy - yz - zx]

= (x + y + z)[x² + y² + z² - xy - yz - zx]

= x³ + y³ +z³ - 3xyz [∵ x³ + y³ +z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)]

Hence, L.H.S = R.H.S

If x + y + z = 0, show that x³ + y³ + z³ = 3xyz

Answer

(x + y + z)³ = x³ + y³ + z³ - 3xyz

Given that x + y + z = 0

= (0)³ = x³ + y³ + z³ - 3xyz

⇒ 3xyz = x³ + y³ + z³

Hence proved

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

Answer

(i) (-12)³ + (7)³ + (5)³

Here,

-12 + 7 + 5 = 0

As, x + y + z = 0

So, x³ + y³ + z³ = 3xyz

= 3 x (-12) x 7 x 5

= -1260

Hence, (-12)³ + (7)³ + (5)³ = -1260

(ii) (28)³ + (-15)³ + (-13)³

Here,

= 28 + (-15) + (-13)

= 28 - 15 - 13

= 0

As, x + y + z = 0

So, x³ + y³ + z³ = 3xyz

= 3 x 28 x (-15) x (-13)

= 16380

Hence, (28)³ + (-15)³ + (-13)³ = 16380

Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:

(i) Area: 25a² - 35a + 12

(ii) Area: 35y² + 13y - 12

Answer

(i) 25a² - 35a + 12

Area of Rectangle = Length x Breadth

= 25a² - 35a + 12

= 25a² -(20a + 15a) + 12

= 25a² -20a -15a + 12

= 5a(5a - 4) - 3(5a - 4)

= (5a - 4)(5a - 3)

Hence, one possible answer is : Length = 5a - 3, Breadth = 5a - 4

(ii) 35y² + 13y - 12

Area of Rectangle = Length x Breadth

= 35y² + 13y - 12

= 35y² +(28y - 15y) - 12

= 35y² + 28y - 15y - 12

= 7y(5y + 4) - 3(5y + 4)

= (5y + 4)(7y - 3)

Hence, one possible answer is : Length = 7y - 3, Breadth = 5y + 4

What are the possible expressions for the dimensions of the cuboids whose volume are given below?

(i) Volume: 3x² - 12x

(ii) Volume: 12ky² + 8ky - 20k

Answer

(i) Volume: 3x² - 12x

Volume of Cuboid = Length x Breadth X Height

= 3x² - 12x

= 3x(x - 4)

= 3(x)(x - 4)

Hence, one possible answer is : 3, x and x - 4.

(ii) Volume: 12ky² + 8ky - 20k

= 12ky² + 8ky - 20k

= 4k(3y² + 2y - 5)

= 4k(3y² + 5y - 3y - 5)

= 4k[y(3y + 5) -1(3y + 5)]

= 4k (y - 1)(3y + 5)

Hence, one possible answer is : 4k, 3y + 5 and y - 1.