Polynomials MCQ Quiz - Objective Question with Answer for Polynomials - Download Free PDF

Given:

p(x) = 2x³ – 3x² – 2x + 3 and q(x) = 3x² + 8x + 5

Concept:

H.C.F. of two or more equations is the greatest factor that divides each of them exactly.

Calculation:

The factors of p(x) = 2x³ – 3x² – 2x + 3

⇒ x² × (2x – 3) – 1 × (2x – 3)

⇒ (x² – 1) × (2x – 3)

⇒ (x – 1) × (x + 1) × (2x – 3)

And, the factors of q(x) = 3x² + 8x + 5

⇒ 3x² + 5x + 3x + 5

⇒ x × (3x + 5) + 1 × (3x + 5)

⇒ (3x + 5) × (x + 1)

Given:

p(x) = x3 – 2x2 – 4x – 1, g(x) = x + 1

Concept Used:

By remainder theorem, the remainder when p(x) divide by g(x')  is p(x')

Calculation:

p(x) = x3 – 2x2 – 4x – 1

g(x) = x + 1

x + 1 = 0 ⇒ x = -1

∴ According to remainder theorem, 

p(-1) = (-1)³ - 2(-1)² - 4(-1) - 1 = -1 - 2 + 4 - 1 = 0

∴ When p(x) is divided by g(x), then remainder is 0.

Given:

Expression 1: (x+3)(2x² - 3x + a)

Expression 2: (x-2)(3x² + 10x - b)

HCF of Expression 1 and Expression 2 is x² + x - 6

Formula used:

If an expression is the HCF of two polynomials, then it must be a factor of both polynomials.

This means that the roots of the HCF polynomial must also be roots of the two given expressions.

Calculation:

First, factorize the HCF: x² + x - 6

⇒ x² + 3x - 2x - 6

⇒ x(x + 3) - 2(x + 3)

⇒ (x + 3)(x - 2)

Since (x + 3)(x - 2) is the HCF, it must be a factor of both given expressions.

Consider Expression 1: (x + 3)(2x² - 3x + a)

We already have the factor (x + 3). For the HCF to be (x + 3)(x - 2), it means that (x - 2) must be a factor of (2x² - 3x + a).

If (x - 2) is a factor of (2x² - 3x + a), then substituting x = 2 into (2x² - 3x + a) should result in 0.

⇒ 2(2)² - 3(2) + a = 0

⇒ 2(4) - 6 + a = 0

⇒ 8 - 6 + a = 0

⇒ 2 + a = 0

⇒ a = -2

Consider Expression 2: (x - 2)(3x² + 10x - b)

We already have the factor (x - 2). For the HCF to be (x + 3)(x - 2), it means that (x + 3) must be a factor of (3x² + 10x - b).

If (x + 3) is a factor of (3x² + 10x - b), then substituting x = -3 into (3x² + 10x - b) should result in 0.

⇒ 3(-3)² + 10(-3) - b = 0

⇒ 3(9) - 30 - b = 0

⇒ 27 - 30 - b = 0

⇒ -3 - b = 0

⇒ b = -3

Now, we need to find the value of (2a - 3b):

⇒ 2a - 3b = 2(-2) - 3(-3)

⇒ 2a - 3b = -4 - (-9)

⇒ 2a - 3b = -4 + 9

⇒ 2a - 3b = 5

∴ The correct answer is option 3.

Given:

The expression to factorize is 6x² - 24xy + 17x + 24y² - 34y + 5

Calculation:

Let's try to group terms or use a systematic approach.

Consider the quadratic terms: 6x² - 24xy + 24y²

We can factor out 6: 6(x² - 4xy + 4y²)

Recognize the perfect square trinomial: 6(x - 2y)²

So the expression becomes: 6(x - 2y)² + 17x - 34y + 5

Notice that 17x - 34y can be factored as 17(x - 2y).

Let P = (x - 2y).

The expression now becomes: 6P² + 17P + 5

This is a quadratic expression in P. We can factor this quadratic.

We need two numbers that multiply to 6 × 5 = 30 and add up to 17. These numbers are 15 and 2.

6P² + 15P + 2P + 5

(6P² + 15P) + (2P + 5)

3P(2P + 5) + 1(2P + 5)

(3P + 1)(2P + 5)

Now substitute P = (x - 2y) back into the factored expression:

[3(x - 2y) + 1][2(x - 2y) + 5]

[3x - 6y + 1][2x - 4y + 5]

So, the factors are (3x - 6y + 1) and (2x - 4y + 5).

∴ The correct answer is option 4.

Concept Used:

According to the remainder theorem, 

When we divide a polynomial P(x) by (x - c), the remainder is P(c)

Calculation:

Let P(x) = 2x3- 9x2 + x + 12 be the given polynomial.

Now,

P(x) = 2x3- 9x2 + x + 12

Using remainder theorem

P(1) = 2 × 1³ - 9 × 1² + 1 + 12

⇒ P(1) = 15 - 9 = 6

Therefore, 6 is the remainder.

Given

2x⁵ + 2x³y³ + 4y⁴ + 5.

Concept

The degree of a polynomial is the highest of the degrees of its individual terms with non-zero coefficients.

Solution

Degree of the polynomial in 2x⁵ = 5

Degree of the polynomial in 2x³y³ = 6

Degree of the polynomial in 4y⁴ = 4

Degree of the polynomial in 5 = 0

Hence, the highest degree is 6

∴ Degree of polynomial = 6

Mistake Points  

One may choose 5 as the correct option due to x⁵ but the correct answer will be 6 as 2x³y³ has the highest power of 6.

Important Points

 The degree of a polynomial is the highest of the degrees of its individual terms with non-zero coefficients. Here for a specific value when x will be equal to y then the equation will be:

2x5 + 2x3y3 + 4y4 + 5

= 2x⁵ + 2x⁶ + 4x⁴ + 5

∴ The degree of the polynomial will be 6

Concept:

If α and β are the zeros of polynomial p(x) then,

p(α) = 0 & p(β) = 0  

Calculation:

Let p(x) =  (k - 1)x2 + kx +1

According to question, x = -3 is one of its zeros, than

p(x) at x = -3 become zero.

Therefore,

(k - 1)(-3)2 + k(-3) +1 = 0

⇒ 9k - 9 - 3k + 1 = 0

⇒ 6k = 8

⇒ k = 4/3

Hence, option 2 is correct.

Shortcut TrickUsing formula a² - b² = (a + b) (a - b)

We can write (x2 + y2 - z2)2 - (x2 - y2 + z2)2 as

(x2 + y2 - z2 + x2 - y2 + z2) (x2 + y2 - z2 - x2 + y2 - z2)

⇒ 2x2 (2y2 - 2z2)

⇒ 4x2y2 - 4x2z2

Alternate Method 

Formula Used:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Calculation:

Let’s put a = x², b = -y², c = z²

⇒ (x² + y² - z²)² = x⁴ + y⁴ + z⁴ + 2x²y² – 2y²z² – 2z²x²      ----(1)

Let’s put a = x², b = y², c = -z²

⇒ (x² - y² + z²)² = x⁴ + y⁴ + z⁴ – 2x²y² – 2y²z² + 2z²x²      ----(2)

(1) – (2)

⇒ 4x²y² – 4z²x²

∴ The require answer is 4x²y² – 4x²z²

Alternate Method

Let x = 1, y = 2 and z = -3

Now put put these value in (x2 + y2 - z2)2 - (x2 - y2 + z2)2 

(1 + 4 - 9)² - (1 - 4 + 9)²

16 - 36 = - 20

Now put x = 1, y = 2 and z = -3 in option

1)  4x2y2 - 4x2z² = 4(1)(2)² - 4(1)(3)² = 16 - 36 = -20

Hence option 1 is correct option

Given

4x⁴ + 3x³ + 2x² + x + 1

Concept

The degree of a polynomial is the highest of the degrees of its individual terms with non-zero coefficients.

Solution

Degree of the polynomial in 4x⁴ = 4

Degree of the polynomial in 3x³ = 3

Degree of the polynomial in 2x² = 2

Degree of the polynomial in x = 1

Hence, the highest degree is 4.

∴ Degree of polynomial = 4

Given:

5x + 3y = 15 and 2xy = 6

Formula used:

(a - b)² = (a + b)² - 4ab

Calculation:

(5x - 3y)² =  (5x + 3y)² - 4 × 5x.3y

⇒ 15² - 30 × 2xy

⇒ 225 - 180 = 45

(5x - 3y) = √45

⇒ \(3\sqrt5\)

∴ The correct option is 2

Given:

The given equation is x⁴ + 2x³ - 16x² - 22x + 7 = 0

One of the root is (2 + \(√3\))

Concept: If all coefficients of the equation are real, then irrational roots will occur in conjugate pairs.

Calculation:

The one root is (2 + \(√3\)) and the conjugate is (2 - \(√3\))

So, the other root is (2 - \(√3\))

⇒ α = 2 + \(\sqrt 3\) and β = 2 - \(√3\)

Product of these roots 

⇒ \((x - 2 - √3) (x - 2 + √3)\)

⇒ (x - 2)² - 3 

⇒ x² - 4x + 1

On dividing x4 + 2x3 - 16x2 - 22x + 7 by x2 - 4x + 1

Then the other quadratic factor is x² + 6x + 7

Then the given equation reduce in the form 

⇒ (x² - 4x + 1)(x² + 6x + 7) = 0

The roots of the equation x² + 6x + 7 = 0

⇒ x = \(\frac{- 6 \ ±\ √{36\ -\ 28}}{2}\)

⇒ x = - 3 ± \(\sqrt2\)

The other root is - 3 ± \(\sqrt2\)

∴ The other root of x4 + 2x3 - 16x2 - 22x + 7 = 0 is - 3 - \(\sqrt2\)

Given:

a = 3 + 2√2

Concept Used:

a² – b² = (a – b)(a + b)

a³ + b³ = (a + b)³ – 3ab(a + b)

Calculation:

a = 3 + 2√2

1/a = 1/(3 + 2√2)

⇒ 1/a = (3 – 2√2)/{(3 + 2√2) × (3 – 2√2)}

⇒ 1/a = (3 – 2√2)/{3² – (2√2)²}

⇒ 1/a = (3 – 2√2)/(9 – 8)

⇒ 1/a = (3 – 2√2)

Now,

a + 1/a = 3 + 2√2 + 3 – 2√2

⇒ a + 1/a = 6

(a⁶ – a⁴ – a² + 1)/a³

⇒ a³ – a – 1/a + 1/a³

⇒ (a³ + 1/a³) – (a + 1/a)

⇒ {(a + 1/a)³ – 3(a + 1/a)} – (a + 1/a)

⇒ (6³ – 3 × 6) – 6

⇒ 216 – 18 – 6

⇒ 192

∴ The required value of (a⁶ – a⁴ – a² + 1)/a³ is 192

Formula used:

(a + b)² = a² + b² + 2ab

(a + b)(a - b) = a² - b²

Calculation:

x⁴ + x² + 25

It can be written as (x²)² + 2 × x² × 5 + (5)² - (3x)²

⇒ x4 + 10x2 + 25 - 9x2

⇒ (x² + 5)² - (3x)²

⇒ (x² + 5 + 3x)(x² + 5 - 3x)

∴ The factors of x⁴ + x² + 25 are (x2 + 3x + 5) (x2 - 3x + 5).

We know that

x³ + y³ = (x + y)(x² + y² – xy)

Now we have x³ + y³ = 22 and x + y = 5

⇒ 22 = 5(x² + y² – xy)

⇒ 22 = 5[(x + y)² − 3xy)]

⇒ 22 = 5[(5)² − 3xy)]

⇒ xy = 103/15

Now multiply x³ + y³ = 22 with x + y = 5

⇒ x⁴ + y⁴ + xy(x² + y²) = 110

⇒ x⁴ + y⁴ = 110 – xy{(x² + y² − 2xy + 2xy)}

⇒ x⁴ + y⁴ = 110 – xy{(x + y)² − 2xy}

xy = 103/15 and x + y = 5

⇒ x⁴ + y⁴ = 110 – 103/15{(5)² − 2 × 103/15}

⇒ x⁴ + y⁴ = 110 – 6.87{(25 –  13.73}

⇒ x4 + y4 = 110 – 6.87 {(11.27)}

⇒ x⁴ + y⁴ = 110 – 77.42

⇒ x4 + y4 = 32.58

∴ Value of x⁴ + y⁴ is 33.

x3 + y3 = (x + y)(x2 + y2 – xy)

⇒ 22 = 5(x2 + y2 – xy)

⇒ 22 = 5[(x + y)2 − 3xy)]

⇒ 22 = 5[(5)2 − 3xy)]

⇒ xy = 103/15

(x³ + y³) (x + y) = x⁴ + y⁴ + xy(x² + y²)

(x³ + y³) (x + y)= (x⁴ + y⁴) + {xy[(x + y)² – 2xy)]

⇒ 22 × 5 = x⁴  + y⁴  + 103/15[25 - 206/15]

⇒ x⁴ + y⁴ = 32.63 ≈ 33

Concept used:

Remainder theorem: 

If a polynomial p(x) is divided by (x−a), then the remainder is a

constant given by p(a).

Calculation:

Let p(x) = 5x3 + 5x2 – 6x + 9 

Since, (x + 3) divide p(x), then, remainder will be p(-3).

⇒ p(-3) = 5 × (-3)3 + 5 × (-3)2 – 6 × (-3) + 9

⇒ p(-3) = -63