Practice Test: Polynomials - Grade 10 MCQ

product of zero =  constant term / coefficient of x²
                         =  d/b 

Given, the quadratic polynomial is (k-1) x² + k x + 1.

One zero of the polynomial is -3.

Explanation:

Let f(x) = (k-1) x² + k x + 1

f(-3) = 0

Put x = -3 in the given polynomial

(k-1) (-3)² + k (-3) + 1 = 0

(k-1)(9) - 3k + 1 = 0

9k - 9 - 3k + 1 = 0

By grouping,

9k - 3k - 9 + 1 = 0

6k - 8 = 0

6k = 8

k = 8/6

k = 4/3

Therefore, the value of k is 4/3.

⇒ p(x) = x²−3x−10
But we know that, if we multiply or divide any polynomial by any arbitrary constant. Then, the zeroes of the polynomial never change.
Therefore,

p(x)=kx²−3kx−10

 Where,k is a real number

Hence, the required number of polynomials having zeros −2 and 5 are infinite i.e. more than 3.

Answer: (a) 3x²-3√2x+1

Explanation: Sum of zeroes = α + β =√2

Product of zeroes = α β = 1/3

∴ If α and β are zeroes of any quadratic polynomial, then the polynomial is;

x²–(α+β)x +αβ

= x² –(√2)x + (1/3)

= 3x²-3√2x+1

Let p(x) =ax³ + bx² + cx + d
Given that, one of the zeroes of the cubic polynomial p(x) is zero.
Let α, β and γ are the zeroes of cubic polynomial p(x), where a = 0.
We know that,

Explanation:

Given Information:
- Two zeros of the cubic polynomial are 0.
- Therefore, the cubic polynomial can be written as ax^3 + bx^2 + cx + d = a(x - 0)(x - 0)(x - k) = ax^3 - akx^2 = ax^3.

Finding the value of c:
- Since the coefficient of the linear term in the polynomial is c, and two of the zeros are given to be 0, the sum of the zeros is 0 + 0 + k = -b/a.
- As k represents the remaining zero, this implies that k = -b/a.
- Since the sum of the zeros of a cubic polynomial is -b/a, the value of c cannot be determined from the given information.
- Therefore, the correct answer is c.​​​​

Here p(3) = 0
⇒ 4(3)² - 6 x 3 - m = 0
⇒ 36 - 18 - m = 0 ⇒ m =18
∴ Value of m is exact divisior of 36.

x² - (sum of roots)x + product of roots
x² - (5+(-8))x + 5x-8
x²​​​​​​​ - (-3)x + -40
x²​​​​​​​ + 3x -40
 

x - 2 is a factor of x³ - 3x² + 4x - 4
∵ remainder is zero
Similarly x + 1 is a factor of 2x³ + 4x + 6
but x - 1 is not a factor of x⁵ + x⁴ - x³ + x² - x + 1
∵ remainder is not zero
∴ Statements 1 and 2 are correct.

as degree of divisor is 4 which is greater than dividend so reaminder would itself be dividend as division cannot occur so degree of remainder would be 3