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

Given:

The polynomial equation is 4x³ - 8x² - 13x + 19 = 0.

Formula used:

Sum of roots = -b/a

Product of all roots = (-1)ⁿ × constant term/a, where n is the degree of the polynomial.

Calculations:

For the polynomial 4x³ - 8x² - 13x + 19 = 0:

Coefficient of x³ (a) = 4

Coefficient of x² (b) = -8

Constant term = 19

Sum of roots:

⇒ Sum of roots = -b/a

⇒ Sum of roots = -(-8)/4 = 2

Product of all roots:

⇒ Product of all roots = (-1)³ × constant term/a

⇒ Product of all roots = -19/4

∴ Sum and product of all roots are 2 and -19/4 respectively.

The correct answer is option 2.

Given:

Polynomial: x⁴ − 10x² + 22

Formula used:

Quadratic formula: For an equation ax² + bx + c = 0, x =

Calculations:

Let y = x². The given polynomial can be written as a quadratic equation in y:

y² - 10y + 22 = 0

Using the quadratic formula to find the roots for y, where a = 1, b = -10, c = 22:

y =

⇒ y =

⇒ y =

⇒ y =

⇒ y = 5

So, the two roots for y are:

y₁ = 5 +

y₂ = 5 -

Therefore, the quadratic in y can be factored as:

(y - y₁)(y - y₂) = (y - (5 + ))(y - (5 - ))

Substitute back y = x²:

(x² - (5 + ))(x² - (5 - ))

∴ The factorization of the polynomial x⁴ − 10x² + 22 into a product of two quadratic polynomials is (x² - (5 + ))(x² - (5 - )).

Given:

Polynomial expressions: (x² - x + 4) and (x³ - 2x² + 3x + 1)

Concept used:

To find the sum of the coefficients of x⁴ and x² in the product of two polynomials, we need to multiply the corresponding terms in each polynomial and then sum the coefficients.

Calculation:

Let's multiply the two polynomials:

⇒ (x² - x + 4) * (x³ - 2x² + 3x + 1)

⇒ x² * x³ + x² * (-2x²) + x² * 3x + x² * 1 - x * x³ - x * (-2x²) + x * 3x + x * 1 + 4 * x³ + 4 * (-2x²) + 4 * 3x + 4 * 1

⇒ x⁵ - 2x⁴ + 3x³ + x² - x⁴ + 2x³ - 3x² + x + 4x³ - 8x² + 12x + 4

⇒ x⁵ - 3x⁴ + 9x³ - 10x² + 13x + 4

The coefficient of x⁴ is -3, and the coefficient of x² is -10.

Therefore, the sum of the coefficients of x⁴ and x² is -3 + (-10) = -13.

Hence, the correct option is -13

Given:

 and q = 3xy - 5x3y4

Calculation:

pq + 2x3y2 

⇒ 

⇒ 

⇒ 

∴ The correct answer is 

Calculation:

Sum of the given expressions,

(5x - 7y + 2) + (3y - 4x + 2xy) + ( 2x - 3xy - 5)

⇒ (5x - 4x + 2x) + (- 7y + 3y) + (2xy - 3xy) + (2 - 5)

⇒  3x - 4y - xy - 3

Here, the coefficients of x, y and xy are 3, - 4, - 1 respectively so p = 3, q = - 4 and r = - 1.

Then the value of p + q - r = 3 + (-4) - (-1) = 3 - 4 + 1 = 0

∴ The correct answer is 0

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

⇒ 

∴ The correct option is 2

Given:

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

One of the root is (2 + )

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

Calculation:

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

So, the other root is (2 - )

⇒ α = 2 +  and β = 2 - 

Product of these roots 

⇒ 

⇒ (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 = 

⇒ x = - 3 ± 

The other root is - 3 ± 

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

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