Explore topic-wise InterviewSolutions in .

(c) (x – y – z)

x² (y – z) + y² (z – x) – z(xy – yz – zx) 

= x²y – x²z + y²z – y²x – zxy + yz² + z²x 

= xy(x – y – z) + z²(x + y) – z (x² – y²) 

= xy(x – y – z) – z (x + y) (x – y – z) = (x – y – z) (xy – yz – zx) 

Hence, (c) is the correct option.

Since (x – k) is the HCF of (x² + x + 12) and (2x² – kx – 9) 

(x – k) will be a factor of 2x² – kx – 9 

∴ 2.k² – k.k – 9 = 0 ⇒ k² – 9 = 0 ⇒ k = ± 3 

Also, the factors of (x² + x – 12) = (x + 4) (x – 3) ∴ k = 3.

(i) Given, polynomial is

p(x) = 10x – 4x² – 3

On putting x = 0,1 and – 2, respectively in Eq. (i),

we get p(0) = 10(0)-4(0)² -3 

= 0-0-3 

= -3

p(1) = 10 (1) - 4 1)² -3

= 10-4-3

= 10-7

= 3

and p(-2) =10 (-2)- 4 (-2)² – 3

= -20-4×4-3 =-20-16-3=-39

Hence, the values of p(0), p(1) and p(-2) are respectively, -3,3 and – 39.

(ii) Given, polynomial is p(y) = (y+2)(y-2)

On putting y =0,1 and -2, respectively in Eq. (i), we get p(0) =(0+2)(0-2)= -4

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

= 3 x (-1) = -3

and p(-2) = (-2 + 2)(-2 -2)

=0 (-4) = 0

Hence, the values of p(0),p(1) and p(-2) are respectively,-4,-3 and 0.

Given, 

q(y) = 7y² – (11/3)y – 2/3 

We put q(y) = 0 

⇒ 7y² – (11/3)y – 2/3 = 0 

⇒  (21y² – 11y -2)/3 = 0 

⇒ 21y² – 11y – 2 = 0 

⇒ 21y² – 14y + 3y – 2 = 0

 ⇒ 7y(3y – 2) – 1(3y + 2) = 0 

⇒ (3y – 2)(7y + 1) = 0 

This gives us 2 zeros, for 

y = 2/3 and y = -1/7 

Hence, the zeros of the quadratic equation are 2/3 and -1/7. 

Now, for verification 

Sum of zeros = – coefficient of y / coefficient of y² 

2/3 + (-1/7) = – (-11/3) / 7 

-11/21 = -11/21 

Product of roots = constant / coefficient of y² 

2/3 x (-1/7) = (-2/3) / 7 

– 2/21 = -2/21 

Therefore, the relationship between zeros and their coefficients is verified.

(d) 2

Remainder when x¹³ + 1 is divided by (x – 1) = 1¹³ + 1 = 2.

(b) 6

Let the cubic polynomial be : 

f(x) = ax3 + bx2 + cx + d. 

Given, f(1) = 1 ⇒ a + b + c + d = 1            ....(i) 

f(2) = 2 ⇒ 8a + 4b + 2c + d = 2                ...(ii) 

f(3) = 4 ⇒ 27a + 9b + 3c + d = 3             ...(iii) 

f(4) = 5 ⇒ 125a + 25b + 5c + d = 5        ...(iv) 

(ii) – (i) ⇒ 7a + 3b + c = 1                       ...(v) 

(iii) – (ii) ⇒ 19a + 5b + c = 1                  ...(vi) 

(iv) – (iii) ⇒ 98a + 16b + 2c = 2           ...(vii) 

(vi) – (v) ⇒ 12a + 2b = 0 ⇒ 6a + b = 0           ...(viii) 

(vii) – 2 (vi) ⇒ 60a + 6b = 0 ⇒ 10a + b = 0           ...(ix) 

Solving (viii) and (ix), we get a = 0 ⇒ b = 0 

Putting a = 0, b = 0 in (v), we, get c = 1 

Also from (i), a = 0, b = 0, c = 1 ⇒ d = 0.

Putting values of a, b, c, d in f(x) = ax³ + bx² + cx + d, we get the polynomial f(x) = x ⇒ f(6) = 6.

Answer is:

(a) for all values of n 

Given, 

p(y) = y² + (3√5/2)y – 5 

We put f(v) = 0 

⇒ y² + (3√5/2)y – 5 = 0 

⇒  y² – √5/2 y + 2√5y – 5 = 0 

⇒ y(y – √5/2) + 2√5 (y – √5/2) = 0 

⇒ (y + 2√5)(y – √5/2) = 0 

This gives us 2 zeros, for 

y = √5/2 and y = -2√5 

Hence, the zeros of the quadratic equation are √5/2 and -2√5. 

Now, for verification 

Sum of zeros = – coefficient of y / coefficient of y² 

√5/2 + (-2√5) = – (3√5/2) / 1 

-3√5/2 = -3√5/2 

Product of roots = constant / coefficient of y² 

√5/2 x (-2√5) = (-5) / 1 

– (√5)² = -5 

-5 = -5 

Therefore, the relationship between zeros and their coefficients is verified.

Given, 

f(x) = x² – (√3 + 1)x + √3 

We put f(x) = 0 

⇒ x² – (√3 + 1)x + √3 = 0 

⇒  x² – √3x – x + √3 = 0 

⇒ x(x – √3) – 1 (x – √3) = 0 

⇒ (x – √3)(x – 1) = 0 

This gives us 2 zeros, for 

x = √3 and x = 1 

Hence, the zeros of the quadratic equation are √3 and 1. 

Now, for verification 

Sum of zeros = – coefficient of x / coefficient of x² 

√3 + 1 = – (-(√3 +1)) / 1 

√3 + 1 = √3 +1 

Product of roots = constant / coefficient of x² 

1 x √3 = √3 / 1 

√3 = √3 

Therefore, the relationship between zeros and their coefficients is verified.

(d) 0

Given, x + y + z = 0 ⇒ x + y = – z 

⇒ x² + y² + 2xy = z² ⇒ x² + y² = z² – 2xy

∴ \(\frac{1}{x^2+y^2-z^2} = \frac{1}{z^2-2xy-z^2}=\frac{1}{-2xy}=-\frac{1}{2xy}\)

Similarly, \(\frac{1}{y^2+z^2-x^2} = -\frac{1}{2xy}\) and \(\frac{1}{z^2+x^2-y^2}=-\frac{1}{2zx}\)

∴ \(\frac{1}{x^2+y^2-z^2}+\frac{1}{y^2+z^2-x^2}+\frac{1}{z^2+x^2-y^2}\)

= \(-\frac{1}{2xy}-\frac{1}{2yz}-\frac{1}{2zx}\) = \(-\frac{1}{2}\big[\frac{z+x+y}{xyz}\big]\) = 0.                   [∵ x + y + z = 0]

Given, 

g(x) = a(x²+1) – x(a²+1) 

We put g(x) = 0 

⇒ a(x²+1)–x(a²+1) = 0 

⇒ ax² + a − a²x – x = 0 

⇒ ax² − a²x – x + a = 0 

⇒ ax(x − a) − 1(x – a) = 0 

⇒ (x – a)(ax – 1) = 0 

This gives us 2 zeros, for 

x = a and x = 1/a 

Hence, the zeros of the quadratic equation are a and 1/a. 

Now, for verification 

Sum of zeros = – coefficient of x / coefficient of x² 

a + 1/a = – (-(a² + 1)) / a 

(a² + 1)/a = (a² + 1)/a 

Product of roots = constant / coefficient of x² 

a x 1/a = a / a 

1 = 1 

Therefore, the relationship between zeros and their coefficients is verified.

Given, 

f(v) = v² + 4√3v – 15 

We put f(v) = 0 

⇒ v² + 4√3v – 15 = 0 

⇒  v² + 5√3v – √3v – 15 = 0 

⇒ v(v + 5√3) – √3 (v + 5√3) = 0 

⇒ (v – √3)(v + 5√3) = 0 

This gives us 2 zeros, for 

v = √3 and v = -5√3 

Hence, the zeros of the quadratic equation are √3 and -5√3. 

Now, for verification 

Sum of zeros = – coefficient of v / coefficient of v² 

√3 + (-5√3) = – (4√3) / 1 

-4√3 = -4√3 

Product of roots = constant / coefficient of v² 

√3 x (-5√3) = (-15) / 1 

-5 x 3 = -15 

-15 = -15 

Therefore, the relationship between zeros and their coefficients is verified.

(d) p⁶ – 6p⁴ + 9p² – 2

Given, \(x+\frac{1}{x}\) = p          ⇒ \(\bigg(x+\frac{1}{x}\bigg)^2\) = p²

⇒ \(x^2+\frac{1}{x^2}+2 = p^2\)    ⇒ \(x^2+\frac{1}{x^2} = p^2 - 2\)

⇒ \(\bigg(x^2+\frac{1}{x^2}\bigg)^3\) = (p² - 2)³

⇒  \(x^6+\frac{1}{x^6}\) + 3\(\bigg(x^2+\frac{1}{x^2}\bigg)\) = p⁶ - 8 + 6p² (p² - 2)

⇒ \(x^6+\frac{1}{x^6}\) + 3 (p² - 2) = p⁶ - 8 + 6p² (p² - 2)

⇒ \(x^6+\frac{1}{x^6}\) = p⁶ - 6p⁴ - 9p² - 2

Given, 

h(s) = 2s² – (1 + 2√2)s + √2 

We put h(s) = 0 

⇒ 2s² – (1 + 2√2)s + √2 = 0 

⇒  2s² – 2√2s – s + √2 = 0 

⇒ 2s(s – √2) -1(s – √2) = 0 

⇒ (2s – 1)(s – √2) = 0 

This gives us 2 zeros, for 

x = √2 and x = 1/2 

Hence, the zeros of the quadratic equation are √3 and 1. 

Now, for verification 

Sum of zeros = – coefficient of s / coefficient of s² 

√2 + 1/2 = – (-(1 + 2√2)) / 2 

(2√2 + 1)/2 = (2√2 +1)/2 

Product of roots = constant / coefficient of s² 

1/2 x √2 = √2 / 2 

√2 / 2 = √2 / 2 

Therefore, the relationship between zeros and their coefficients is verified.

Let P and Q be the two expressions. Then, 

P + Q = 5a² – a – 4                            ...(i) 

P – Q = a² + 9a – 10                         ...(ii) 

Adding (i) and (ii) 

⇒ 2P = 6a² + 8a – 14 ⇒ P = 3a² + 4a – 7 = (a – 1) (3a + 7) 

From (i), Q = (5a² – a – 4) – (3a² + 4a – 7) = 2a² – 5a + 3 = (a – 1) (2a – 3)

∴ LCM of P and Q = (a – 1) (2a – 3) (3a + 7).

(d) \(\frac{a}{3}.\)

a^x . a^y . a^z = (x + y + z)^{x + y + z} 

⇒ a^{x + y + z} = (x + y + z)^{x + y + z} 

⇒ a = (x + y + z) 

Now, (x + y + z)^y = a^x (given) 

⇒ (x + y + z)^y = (x + y + z)^x ⇒ y = x 

Similarly, y = z and z = x.

∴ x = y = z = \(\frac{x+y+z}{3}\) = \(\frac{a}{3}.\)

\(\frac{1}{y+z}+\frac{1}{z+x} = \frac{2}{x+y}\)

⇒ \(\frac{1}{y+z}+\frac{1}{x+z} \) = \(\frac{1}{x+y}+\frac{1}{z+x} \) ⇒ \(\frac{(x+y)-(y+z)}{(y+z)(x+y)}\) = \(\frac{(z+x)-(x+y)}{(x+y)(z+x)}\)

⇒ \(\frac{x-z}{y+z} = \frac{z-y}{z+x}\) ⇒ (x-z)(x+z) = (z-y)(z+y)

⇒ x² – z² = z² – y² ⇒ x² + y² = 2z².

(c) a³ + a²  – 3a – 2

Given, \(x+\frac{1}{x}=a\) 

Now,  x³ + x² + \(\frac{1}{x^3}+\frac{1}{x^2}\) = \(\big(x^3+\frac{1}{x^3}\big)\)+ \(\big(x^2+\frac{1}{x^2}\big)\)

= \(\big(x+\frac{1}{x}\big)^3\) - 3\(\big(x+\frac{1}{x}\big)\) + \(\big(x+\frac{1}{x}\big)^2\) - 2

= a³ – 3a + a² – 2 = a³ + a² – 3a – 2.

x = \(\frac{a-b}{a+b}\) ⇒ \(\frac{1}{x}\) = \(\frac{a+b}{a-b}\)

⇒ \(\frac{1+x}{1-x}\) = \(\frac{a+b+a-b}{a+b-a+b}\) = \(\frac{2a}{2b}\) ⇒ \(\frac{1+x}{1-x}\) = \(\frac{a}{b}\)          (Applying componendo and dividendo)

Similarly, \(\frac{1+y}{1-y}\) = \(\frac{b}{c}\), \(\frac{1+z}{1-z}\) = \(\frac{c}{a}\) ∴ \(\frac{1+x}{1-x}\) . \(\frac{1+y}{1-y}\) . \(\frac{1+z}{1-z}\) = \(\frac{a}{b}\).\(\frac{b}{c}\).\(\frac{c}{a}\) = 1.

(c) 0

Let \(\frac{x}{(b-c)(b+c-2a)}=\frac{y}{(c-a)(c+a-2b)}=\frac{z}{(a-b)(a+b-2c)}\) = k.

Then, x = k(b – c) (b + c – 2a) 

y = k(c – a) (c + a – 2b) 

z = k(a – b) (a + b – 2c) 

∴ x + y + z = k(b – c) (b + c – 2a) + k(c – a) (c + a – 2b) + k(a – b) (a + b – 2c) 

= k(b² – c² – 2ab + 2ca) + k(c² – a² – 2bc + 2ab) + k (a² – b² – 2ca + 2bc) 

= k(b² – c² – 2ab + 2ca + c² – a² – 2bc + 2ab + a² – b² – 2ca + 2bc) 

= k × 0 = 0.

Given, a + b + c = 0. 

Now, a⁴ + b⁴ + c⁴ – 2a²b² – 2b²c² – 2c²a² = (a² + b² + c²)² – 4a²b² – 4b²c² – 4c²a² 

= [(a + b + c)² – 2ab – 2bc – 2ca]² – 4a²b² – 4b²c² – 4c²a² 

= [0² – 2ab – 2bc – 2ca]² – 4a²b² – 4b²c² – 4c²a² 

= 4a²b² + 4b²c² + 4c²a² + 8ab²c + 8abc² + 8a²bc – 4a²b² – 4b²c² – 4c²a² 

= 8ab²c + 8abc² + 8a²bc = 8abc (b + c + a) = 8abc. 0 = 0.

(c) \(\frac{5}2\)

Let the zeroes of the polynomial be α and α + 4 

Here, p(x) = 4x² – 8kx + 9 

Comparing the given polynomial with ax² + bx + c, 

we get: 

a = 4, b = -8k and c = 9

Now, sum of the roots = \(\frac{-b}a\)

⇒ α + α + 4 = \(\frac{-(-8)}4\)

⇒ 2α + 4 = 2k 

⇒ α + 2 = k 

⇒ α = (k – 2) ….(i) 

Also, product of the roots, αβ = \(\frac{c}a\)

⇒ α (α + 4) = \(\frac{9}4\)

⇒ (k – 2) (k – 2 + 4) = \(\frac{9}4\)

⇒ (k – 2) (k + 2) = \(\frac{9}4\)

⇒ k² – 4 = \(\frac{9}4\)

⇒ 4k² – 16 = 9 

⇒ 4k² = 25

⇒ k² = \(\frac{25}4\)

⇒ k = \(\frac{5}4\)

(∵ k >0)

From the question, it’s given that: 

The quadratic polynomial f(x) where a = 4, b = -8k and c = – 9 

And, for roots to be negative of each other, let the roots be α and – α. 

Sum of the roots = α – α = \(\frac{-b}{a}\) 

= – \(\frac{(-8k)}{1}\) 

= 8k = 0 [∵ α – α = 0] 

⇒ k = 0

Here, p(x) = x² + 2x – 195 

Let p(x) = 0 

⇒ x² + (15 – 13)x – 195 = 0 

⇒ x² + 15x – 13x – 195 = 0 

⇒ x (x + 15) – 13(x + 15) = 0 

⇒ (x + 15) (x – 13) = 0 

⇒ x = –15, 13 

Hence, 

the zeroes are –15 and 13.

The correct option is: (C)

Explanation:

For more than three distinct real roots the graph must cut x-axis at least four times. So, graph in option (C) has more than three distinct real roots.

A quadratic equation when sum and product of its zeros is given by:

f(x) = k{x² - (sum of zeros)x + product of the zeros},

where k is a constant

α + β = 24 ....(1)

α - β = 8 ....(2)

Adding 1 and 2 we get,

α + β + α - β = 24 + 8

⇒ 2α = 32

⇒ α = 16

Substitute value in 1 to get

16 + β = 24

⇒ β = 24 -16

⇒ β = 8

α = 16 and β = 8

f(x) = k{x² - (24)x + 16 × 8}

f(x) = k(x² - 24x + 128)

If we will put the different values of k,

we will find the different quadratic equations.

The correct option is: (D) 0

Explanation:

Since, 3 is one of the zeroes of polynomial p(x). So, p(3) = 0

When x³ - 3x² + 3x + 5 is divided by x² - x + 1, the quotient and remainder are x - 2, 7.

From the question, it’s given that: 

α and β are the roots of the quadratic polynomial p(x) where a = 4, b = -5 and c = -1 

Sum of the roots = α+β = \(\frac{-b}{a}\) 

= – \(\frac{(-5)}{4}\)

= \(\frac{5}{4}\) 

Product of the roots = αβ = \(\frac{c}{a}\)

= \(\frac{-1}{4}\) 

α²β + αβ² 

⇒ αβ(α +β) 

⇒ (\(\frac{-1}{4}\))(\(\frac{5}{4}\)) 

= \(\frac{-5}{16}\)

Given, 

The quadratic polynomial f(t) = kt² + 2t + 3k, where a = k, b = 2 and c = 3k. 

And,

Sum of the roots = Product of the roots 

⇒ (-b/a) = (c/a) 

⇒ (-2/k) = (3k/k) 

⇒ (-2/k) = 3 

∴ k = -2/3

Correct option is (A) a = 7, b = -18

\(\because\) (x + 2) and (x – 1) are factors of \(x^3 + 10x^2 + ax + b.\)

\(\Rightarrow\) x = -2 and x = 1 are zeros of \(x^3 + 10x^2 + ax + b.\)

\(\therefore\) \((-2)^3+10.(-2)^2+a\times-2+b=0\) and 1+10+a+b = 0

\(\Rightarrow\) -8+40-2a+b = 0 and b = -a - 11

\(\Rightarrow\) 32 - 2a - a - 11 = 0

\(\Rightarrow\) 3a = 32 - 11 = 21

\(\Rightarrow\) a = \(\frac{21}3\) = 7

\(\therefore\) b = -a - 11 = -7 - 11 = -18

Correct option is A) a = 7,b = -18

Correct option is (A) k = -1

p(3) = q(3)

\(\Rightarrow\) 27k+36+9-4 = 27 - 12 + k

\(\Rightarrow\) 26k + 41 - 15 = 0

\(\Rightarrow\) 26k = 15 - 41 = -26

\(\Rightarrow\) k = \(\frac{-26}{26}\) = -1

Correct option is A) k = -1

Correct option is (C) 2x² – x + 4 = 0

Let \(\alpha\;and\;\beta\) are root of \(2x^2+3x+5=0\)

\(\therefore\) \(\alpha+\beta\) \(=\frac{-3}2\) and \(\alpha\beta\) \(=\frac{5}2\)

Now, \((\alpha+1)+(\beta+1)\) \(=\alpha+\beta+2\) \(=\frac{-3}2+2\) \(=\frac{-3+4}2=\frac12\)

and \((\alpha+1)(\beta+1)\) \(=\alpha\beta+\alpha+\beta+1\) \(=\frac52-\frac32+1\) = 1+1 = 2

\(\therefore\) Equation whose roots are \((\alpha+1)\;and\;(\beta+1)\) is \(x^2-((\alpha+1)+(\beta+1))x+(\alpha+1)(\beta+1)=0\)

\(\Rightarrow\) \(x^2-\frac x2+2=0\)

\(\Rightarrow\) \(2x^2-x+4=0\)

\(\therefore\) Required equation is \(2x^2-x+4=0.\)

Correct option is C) 2x² – x + 4 = 0

(c) 18

If f(x) = 4x² – 6x + m is divisible by (x – 3), then f(3) = 0 

⇒ 4 (3)² – 6.3 + m = 0 ⇒ 36 – 18 + m = 0 ⇒ m = –18. 

The greatest divisor of m = 18.

Let α and β be the zeroes of the required polynomial f(x). 

Then (α + β) = 0 and αβ = -1 

∴f(x) = x² ˗ (α + β)x + αβ

⇒ f(x) = x² ˗ 0x + (-1) 

⇒ f(x) = x² ˗ 1 

Hence, 

required polynomial f(x) = x² ˗ 1. 

∴f(x) = 0 ⇒ x² ˗ 1 = 0 

⇒ (x + 1) (x – 1) = 0 

⇒ (x + 1) = 0 or (x – 1) = 0 

⇒ x = -1 or x = 1 

So, the zeroes of f(x) are -1 and 1.

Let ∝ = 2 and β = -6

Sum of the zeroes = (∝ + β) = 2 – 6 = -4

Product of the zeroes, = 2(-6) = -12

Required quadratic polynomial is

x² – (∝+β)x + ∝β 

= x² – (-4)x – 12

= x² + 4x – 12

And,

Sum of the zeroes = – 4 = -4/1 = (-Coefficient of x)/(Cofficient of x²)

Product of zeroes = -12 = -12/1 = Constant term / Coefficient of x²

Let ∝ = 2/3 and β = -1/4

Sum of the zeroes = (∝ + β) = 2/3 + -1/4 = 5/12

Product of the zeroes, = 2/3 x -1/4 = -1/6

Required quadratic polynomial is

x² – (∝+β)x + ∝β 

= x² – (5/12)x – (-1/6)

= 1/12(12x² – 5x – 2)

And,

Sum of the zeroes = 5/12 = (-Coefficient of x)/(Cofficient of x²)

Product of zeroes = -1/6 = Constant term / Coefficient of x²

Given : Sum of zeroes = (∝ + β) = 8

Product of the zeroes = = 12

Required quadratic polynomial is

x² – (∝+β)x + ∝β 

= x² – (8)x + 12

Now, find the zeroes of the above polynomial.

Let f(x) = x² – (8)x + 12

= x² – 6x – 2x + 12

= (x -6)(x – 2)

Substitute f(x) = 0.

either (x -6) = 0 or (x – 2) = 0

 x = 6 or x = 2

2 and 6 are the zeroes of the polynomial.

We can find the quadratic equation if we know the sum of the roots and product of the roots by using the formula 

x² – (Sum of the roots)x + Product of roots = 0 

⇒ x² – √2x + 1/3 = 0 

⇒ 3x² –3√2x + 1 = 0

Given: 

ax² + 7x + b = 0 

Since, 

x = \(\frac{2}3\) is the root of the above quadratic equation 

Hence, it will satisfy the above equation. 

Therefore, we will get 

a (\(\frac{2}3\))² + 7 (\(\frac{2}3\)) + b = 0 

⇒ \(\frac{4}9\)a + \(\frac{14}3\) + b = 0 

⇒ 4a + 42 + 9b = 0 

⇒ 4a + 9b = – 42 …(1) 

Since, x = –3 is the root of the above quadratic equation 

Hence, It will satisfy the above equation. 

Therefore, we will get 

a (–3)² + 7 (–3) + b = 0 

⇒ 9a – 21 + b = 0 

⇒ 9a + b = 21 …..(2) 

From (1) and (2), we get 

a = 3, b = – 6

The only value of k for which the quadratic polynomial kx² + x + k has equal zeroes is 1/2: False.

A quadratic polynomial kx² + x + k have equal roots when its discriminate is equal to

i.e. D = 0

b² - 4ac = 0

b² = 4ac

1² = 4(k) (k)

1 = 4k²

k² = 1/4

k = +- 1/2

Hence, k = 1/2 is not the only value for which the polynomial has equal roots.

Answer is (B) 18a³b²

36a⁵b² = 3 × 3 × 2 × 2 × a⁵ × b²

90a³b⁴ = 3 × 3 × 2 × 5 × a³ × b⁴ 

= 3 × 3 × 2 × a³ × b²

H.C.F. = 18a³b²

u(x) = (x – 1)²

⇒ u(x) = (x – 1)(x – 1)

and v(x) = (x² – 1) 

= (x – 1)(x + 1)

Product of common least powers 

= (x – 1)

H.C.F. = (x – 1)

Product of highest power of prime factors 

= (x – 1)²(x + 1)

L.C.M. = (x – 1)²(x + 1)

Test:

u(x) × v(x) = (x – 1)² × (x² – 1) 

= (x – 1)²(x² – 1)

and H.C.F. × L.C.M. 

= (x – 1)(x – 1)²(x + 1) 

= (x – 1)²(x² – 1)

So, L.C.M. × H.C.F. = w(x) × v(x).