Factorisation — Competency Focused Questions

One of the factors of (x² - 4x)(x² - 4x - 1) - 20 is :

1. x - 1

2. x - 2

3. x - 4

4. x + 5

Answer

Given,

(x² - 4x)(x² - 4x - 1) - 20

Let us consider y = (x² - 4x)

⇒ (y)(y - 1) - 20

⇒ y² - y - 20

⇒ y² - 5y + 4y - 20

⇒ y(y - 5) + 4(y - 5)

⇒ (y + 4)(y - 5)

⇒ (x² - 4x + 4)(x² - 4x - 5)

⇒ [(x)² - 2(2)(x) + (2)²](x² - 5x + x - 5)

⇒ (x - 2)²[x(x - 5) + 1(x - 5)]

⇒ (x - 2)(x - 2)(x + 1)(x - 5)

Hence, option 2 is correct option.

x⁴ + 7x² + 16 can be factorized as:

1. (x² + x + 4)(x² - x + 4)

2. (x² + x - 4)(x² - x + 4)

3. (x² - x - 4)(x² + x + 4)

4. none of these

Answer

Given,

⇒ x⁴ + 7x² + 16

⇒ x⁴ + 8x² - x² + 16

⇒ x⁴ + 8x² + 16 - x²

⇒ (x²)² + 2(x²)(4) + (4)² - x²

⇒ (x² + 4)² - x²

⇒ (x² + 4 + x)(x² + 4 - x)

⇒ (x² + x + 4)(x² - x + 4).

Hence, option 1 is correct option.

If $x = \dfrac{a + b + c}{3}$, then (x - a)³ + (x - b)³ + (x - c)³ can be factorized as:

1. (x - a)(x - b)(x - c)

2. $\dfrac{(x - a)(x - b)(x - c)}{3}$

3. 3(x - a)(x - b)(x - c)

4. none of these

Answer

Given,

$x = \dfrac{a + b + c}{3}$

Let, p = x - a, q = x - b, r = x - c

Adding,

⇒ p + q + r = (x - a) + (x - b) + (x - c)

⇒ p + q + r = 3x - a - b - c

⇒ p + q + r = 3x - (a + b + c)

⇒ p + q + r = 3 $\Big(\dfrac{a + b + c}{3}\Big)$ - (a + b + c)

⇒ p + q + r = (a + b + c) - (a + b + c)

⇒ p + q + r = 0

If p + q + r = 0, we use identity,

p³ + q³ + r³ = 3pqr

⇒ (x - a)³ + (x - b)³ + (x - c)³ = 3(x - a)(x - b)(x - c).

Hence, option 3 is correct option.

An expression is factorized as (2x³ + 2x² + x)(x² - 2x + 2). Which of the following terms will appear in the simplest form of the above expression?

1. -2x³

2. 3x³

3. 6x³

4. x³

Answer

Given,

⇒ (2x³ + 2x² + x)(x² - 2x + 2)

⇒ 2x³(x² - 2x + 2) + 2x²(x² - 2x + 2) + x(x² - 2x + 2)

⇒ 2x⁵ - 4x⁴ + 4x³ + 2x⁴ - 4x³ + 4x² + x³ - 2x² + 2x

⇒ 2x⁵ - 2x⁴ + 2x² + x³ + 2x

The that term appears in its simplest form is x³

Hence, option 4 is correct option.

If x + y = 7 and x² + y² = 25, then find the value of $\sqrt{xy}$.

Answer

Given,

x + y = 7

x² + y² = 25

By using the identity,

(x + y)² = x² + y² + 2xy

⇒ (7)² = 25 + 2xy

⇒ 49 = 25 + 2xy

⇒ 49 - 25 = 2xy

⇒ 2xy = 24

⇒ xy = $\dfrac{24}{2}$

⇒ xy = 12

⇒ $\sqrt{xy} = \sqrt{12}$

⇒ $\sqrt{xy} = \sqrt{4 × 3}$

⇒ $\sqrt{xy} = 2\sqrt{3}$.

Hence, $\sqrt{xy} = 2\sqrt{3}$.

An expression was factorized as (x - 1)(x - 3)(x - 5) ..... (x - 99). What is the coefficient of x⁴⁹ in the expression ?

Answer

Given,

Polynomial = (x - 1)(x - 3)(x - 5) ........ (x - 99)

Here roots are 1, 3, 5, ....., 99. Let number of roots be n.

The above sequence is in an A.P. with first term (a) = 1, common difference (d) = 2 and last term (a_n) = 99

By formula,

⇒ a_n = a + (n - 1)d

⇒ 99 = 1 + 2(n - 1)

⇒ 99 - 1 = 2(n - 1)

⇒ 98 = 2(n - 1)

⇒ n - 1 = $\dfrac{98}{2}$

⇒ n - 1 = 49

⇒ n = 50.

Sum of roots = $\dfrac{n}{2}(a + l) = \dfrac{50}{2}(1 + 99) = \dfrac{50}{2} \times 100$ = 2500.

For a polynomial of the form,

(a - a₁)(x - a₂)..........

The coefficient of x^{n - 1} is the negative of the sum of all roots.

Thus, the coefficient of x⁴⁹ = -2500.

Hence, coefficient of x⁴⁹ = -2500.

What is the simplified form of the expression

$\dfrac{\dfrac{b^4 - a^4}{a(a + b)} - \dfrac{b^3}{a}}{b^2 - ab + a^2}$?

Answer

Given,

$\dfrac{\dfrac{b^4 - a^4}{a(a + b)} - \dfrac{b^3}{a}}{b^2 - ab + a^2}$

Simplifying the numerator,

$\Rightarrow \dfrac{b^4 - a^4}{a(a + b)} - \dfrac{b^3}{a} \\[1em] \Rightarrow \dfrac{(b^2)^2 - (a^2)^2}{a(a + b)} - \dfrac{b^3}{a} \\[1em] \Rightarrow \dfrac{(b^2 - a^2)(b^2 + a^2)}{a(a + b)} - \dfrac{b^3}{a} \\[1em] \Rightarrow \dfrac{(b - a)(b + a)(b^2 + a^2)}{a(a + b)} - \dfrac{b^3}{a} \\[1em] \Rightarrow \dfrac{(b - a)(b^2 + a^2)}{a} - \dfrac{b^3}{a} \\[1em] \Rightarrow \dfrac{b^3 + ba^2 - ab^2 - a^3}{a} - \dfrac{b^3}{a} \\[1em] \Rightarrow \dfrac{b^3 + ba^2 - ab^2 - a^3 - b^3}{a} \\[1em] \Rightarrow \dfrac{ba^2 - ab^2 - a^3}{a} \\[1em] \Rightarrow \dfrac{a(ba - b^2 - a^2)}{a} \\[1em] \Rightarrow (ab - b^2 - a^2) \\[1em] \Rightarrow -(a^2 + b^2 - ab).$

Substituting value of numerator in given fraction,

$\Rightarrow \dfrac{-(a^2 + b^2 - ab)}{b^2 - ab + a^2} \\[1em] \Rightarrow \dfrac{-(a^2 + b^2 - ab)}{(a^2 + b^2 - ab)} \\[1em] \Rightarrow -1.$

Hence, $\dfrac{\dfrac{b^4 - a^4}{a(a + b)} - \dfrac{b^3}{a}}{b^2 - ab + a^2} = -1$.