Test: Polynomial Division - Grade 10 MCQ

If α is a zero of the polynomial f(x).
The divisor will be x - α.
For example, if 5 is a zero of a polynomial f(x), then its divisor will be x - 5.

We know that,
f(x) = q(x) × g(x) + r(x)
Where, f(x) is the dividend, q(x) is the quotient, g(x) is the divisor and r(x) is the remainder.
∴ 30x⁴ - 50x³ +109x² - 23x + 25 = (10x² + 3)(3x² - 5x + 10) + ax + b
30x⁴ - 50x³ + 109x² - 23x + 25 = 30x⁴-50x³ + 109x²-15x + 30 + ax + b
30x⁴ - 50x³ + 109x² - 23x + 25 - (30x⁴ - 50x³ + 109x² - 15x + 30) = ax + b
-23x + 25 + 15x - 30 = ax + b
-8x - 5 = ax+b
∴ a = -8, b = -5

We know that,
f(x) = q(x) × g(x) + r(x)
Where, f(x) is the dividend, q(x) is the quotient, g(x) is the divisor and r(x) is the remainder.
f(x) = 5x² × (2x+3)+10x + 20
f(x) = 10x³ + 15x² + 10x + 20

On dividing, 15x⁵ + 70x⁴ + 35x³ - 135x² - 40x - 11 by 5x⁴ + 10x³ - 15x² - 5x
We get, 3x + 8 as quotient and remainder as -11.
So if we subtract -11 from 15x⁵ + 70x⁴+35x³ - 135x² - 40x - 11 it will be exactly divisible by 5x⁴ + 10x³ - 15x² - 5x.

x² + kx - 15 is exactly divisible by x - 5
Dividing, x² + kx + 15 by x - 5
We get, 5k + 10 as remainder.
Since, x² + kx - 15 is exactly divisible by 2x - 5
∴ 5k + 10 = 0
k = -2

Consider, f(x) is 27x² - 39x, q(x) as 9x + 2, g(x) as 3x - 5 and remainder is 10.
f(x) = q(x) × g(x) + r(x)
RHS
q(x) × g(x) + r(x) = (9x + 2)(3x - 5) + 10 = 27x² - 45x + 6x - 10 + 10 = 27x² - 39x, which is equal to LHS.
Hence proved.

Since the zeros of the polynomial are 3 and -2.
The divisor of the polynomial will be (x-3) and (x+2).
Multiplying (x - 3) and (x + 2) = x² + 2x - 3x - 6 = x²- x + 6
Dividing, x³ + (6-√3)x² + (-1-√3)x + 30 - 6√3 by x² - x + 6
We get, x - 5 + √3 as quotient.
Hence, the third zero will be 5 - √3.

We know that,
f(x) = q(x) × g(x) + r(x)
Where, f(x) is the dividend, q(x) is the quotient, g(x) is the divisor and r(x) is the remainder.
∴ 50x² - 90x - 25 = q(x) × 5x - 10 - 5
50x² - 90x - 25 + 5 = q(x) × 5x - 10
50x² − 90x − 205x − 10 = q(x)
We get, q(x) = 10x + 2

We know that,
f(x) = q(x) × g(x) + r(x)
Where, f(x) is the dividend, q(x) is the quotient, g(x) is the divisor and r(x) is the remainder.
acx³ + bcx + d = cx × g(x) + d
acx³ + bcx + d – d = cx × g(x)
acx³ + bcxcx = g(x)
g(x) = ax²+b

81x² - 31 is exactly divisible by 9x + 1
Hence, on dividing 81x² - 31 by 9x + 1
We get, 9x - 1 as quotient and remainder as -30.
So if we add 30 to 81x² - 31, it will be exactly divisible by 9x + 1.