Test: Polynomials (Hard) - Class 10 MCQ

Thus, when 6x + 2 is subtracted from the given polynomial 8x⁴ + 14x³ + x² + 7x + 8, then it will be divisible by 4x² - 3x + 2.

Now checking by putting y = 1, 2, 3... until we get solution

y = 1; => x² =15 not possible

y = 2; => x² = 57 not possible

y = 3; => x² = 127 not possible

y = 4; => x² = 225 => x = 15

Hence, the smallest solution is (x, y) = (15, 4).

and , x = , x =

= x² - 5 / 3 or 3x² – 5 is a factor of the given polynomial.

Now, we apply the division algorithm to the given polynomial and 3x² – 5.

So, 3x⁴ + 6x³ – 2x² – 10x – 5 = (3x² – 5) (x² + 2x + 1) + 0

Quotient = x² + 2x + 1 = (x + 1)²

Zeroes of (x + 1)² are –1, –1.

Hence, all its zeroes are

, , -1, -1

using the Quadratic Formula where

a = 4, b = 87, and c = 125

x= −b ± b² − 4ac−−−−−−−√2a

x=−87 ± 872 − 4(4)(125)−−−−−−−−−−−−√2(4)

x= −87 ± 7569 − 2000−−−−−−−−−−√8

x= −87 ± 5569−−−−√8

The discriminant b² − 4ac > 0

so, there are two real roots.

Cannot Simplify the Radical:

x = −87 ± 5569−−−−√8

x = −878 ± 5569−−−−√8

Simplify fractions and/or signs:

x = −878 ± 5569−−−−√8

which becomes x = −1.54678

x = −20.2032

Zeros of the polynomial obtained by taking P(x) = 0

x² + 3x - (a + 2)(a - 1) = 0

x² + (a + 2)x - (a - 1)x - (a + 2)(a - 1) = 0

x[x + a + 2] - (a - 1)[x + a + 2] = 0

[x + a + 2][x - (a - 1)] = 0

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

Hence, the zeros of the polynomial are (a + 2) and (a - 1).

• The value zero" 0" can be considered as a (constant) polynomial, called the zero polynomial.

• Like any other constant value, It has no nonzero terms.

• There we can say that it has no degree either.

• However, its degree is undefined.

Thus a zero is a polynomial of degree zero.

First, a linear polynomial is in the form of ax + b, a ≠ 0, a, b ∈ R

Degree of the polynomial = highest degree of the terms

So, here the highest degree is 1.

Hence, Linear polynomial has only one zero.

For example: A quadratic equation ax² + bx + c = 0 can have 2 zeros, as the highest power of x is 2 or as the degree is 2. ax³ + bx² + cx + d = 0, a cubic equation can have 3 zeros, as the highest power of x is 3 or as the degree is 3.

Given that x = 2

Substitute x in 4x²- 7

4(2²) - 7

4(4) - 7

16 - 7 = 9

3x² - 8x + 2k + 1

a = 3, b = -8 ,c = 2k + 1

suppose the zeroes of the equation are α and β. By the given condition α = 7β

As we know , α + β = -b / a

7β + β = -(-8) / 3

8β = 8 / 3

b = 8 / 3 × 1 / 8 = 1 / 3

αβ = c /a

7β × β = 2k + 1 / 3

7 × 1 / 3 × 1 / 3 = 2k + 1 / 3

7 / 9 = 2k + 1 / 3

21 / 9 = 2k + 1

21 / 9 - 1 = 2k

12 / 9 = 2k

4 / 3 = 2k

4 / 6 = k

2 / 3 = k

f(x) = 0 ⇒ x³ − 5x² − 16x + 80 = 0

⇒ x² (x − 5) − 16(x − 5) = 0

⇒ (x − 5)(x² − 16) = 0

⇒ (x − 5)(x² − 42) = 0

⇒ (x − 5)(x + 4)(x − 4) = 0

⇒ x = 4, −4, 5

∴ The zeroes of the polynomial are 4, −4, 5

Here, p(x) = 6x² + 2x + 2g(x) = 2q(x) = 3x² + x + 1 r(x) = 0

The Degrees of p(x) and q(x) are the same i.e. 2.

Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)

Or, 6x² + 2x + 2 = 2x (3x² + x + 1)

Hence, the division algorithm is satisfied.

Linear term of polynomial implies the coefficient of x, (c in the above equation), and the constant term implies the term independent of x, (d in above equation).

Given, two zeroes of the cubic polynomial are zero.

Let two zeros ie.,β,γ = 0. f(x) = (x − α)(x − β)(x − γ)

f(x) = (x − α)(x − 0)(x − 0)

f(x) = (x − α)(x²)

f(x) = (x³ −x2α)

then the equation does not have a linear term (coefficient of x is 0) and constant term.

Let α, β are the roots of ax² + bx + c

∴ α + β = -b / a and αβ = c / a

Given α and β both are positive.

⇒ b / a is negative.

⇒ a and b both have different signs.

So, the given statement is false.

False, x² −1 cannot be the quotient on division of x6 − 2x³ + x − 1 by a polynomial in degree 5 because the degree of the product of the quotient and the divisor should be equal to the power of the dividend. Here, the degree of the product of the quotient and the divisor is 7. But the degree of x⁶ − 2x³ + x − 1 is 6.

The exponent of the first term is 2.

The exponent of the second term is 1 because bx = bx¹.

The exponent of the third term is 0 because c = cx⁰.

Since the highest exponent is 2, therefore, the degree of ax² + bx + c is 2.

Since, the degree of the polynomial is 2, hence, the polynomial ax² + bx + c can have zero, one or two zeroes.

Hence, the polynomial ax² + bx + c can have at most two zeros.