ACT > Test: Polynomial Division

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Test: Polynomial Division - Question 1

If the polynomial f(x) = x^{2 }+ kx - 15, is exactly divisible by x - 5, then the value of k is _______

Detailed Solution for Test: Polynomial Division - Question 1

x^{2 }+ kx - 15 is exactly divisible by x - 5

Dividing, x^{2 }+ kx + 15 by x - 5

We get, 5k + 10 as remainder.

Since, x^{2 }+ kx - 15 is exactly divisible by 2x - 5

∴ 5k + 10 = 0

k = -2

Test: Polynomial Division - Question 2

The real number that should be subtracted from the polynomial f(x) = 15x^{5 }+ 70x^{4 }+ 35x^{3 }- 135x^{2 }- 40x - 11 so that it is exactly divisible by 5x^{4 }+ 10x^{3 }- 15x^{2 }- 5x is ____________

Detailed Solution for Test: Polynomial Division - Question 2

On dividing, 15x^{5 }+ 70x^{4 }+ 35x^{3 }- 135x^{2 }- 40x - 11 by 5x^{4 }+ 10x^{3 }- 15x^{2 }- 5x

We get, 3x + 8 as quotient and remainder as -11.

So if we subtract -11 from 15x^{5 }+ 70x^{4 }+ 35x^{3 }- 135x^{2 }- 40x - 11 it will be exactly divisible by 5x^{4 }+ 10x^{3 }- 15x^{2 }- 5x.

Test: Polynomial Division - Question 3

The polynomial (x), if the divisor is 5x^{2}, quotient is 2x + 3, and remainder is 10x + 20 is __________

Detailed Solution for Test: Polynomial Division - Question 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.

f(x) = 5x^{2 }× (2x + 3) + 10x + 20

f(x) = 10x^{3 }+ 15x^{2 }+ 10x + 20

Test: Polynomial Division - Question 4

What will be the value of a and b if the polynomial f(x) = 30x^{4 }- 50x^{3 }+ 109x^{2 }- 23x + 25, when divided by 3x^{2 }- 5x + 10, gives 10x^{2 }+ 3 as quotient and ax + b as remainder?

Detailed Solution for Test: Polynomial Division - Question 4

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.

∴ 30x^{4 }- 50x^{3 }+ 109x^{2 }- 23x + 25 = (10x^{2 }+ 3)(3x^{2 }- 5x + 10) + ax + b

30x^{4 }- 50x^{3 }+ 109x^{2 }- 23x + 25 = 30x^{4 }- 50x^{3 }+ 109x^{2 }- 15x + 30 + ax + b

30x^{4 }- 50x^{3 }+ 109x^{2 }- 23x + 25 - (30x^{4 }- 50x^{3 }+ 109x^{2 }- 15x + 30) = ax + b

-23x + 25 + 15x - 30 = ax + b

-8x - 5 = ax + b

∴ a = -8, b = -5

Test: Polynomial Division - Question 5

If α is a zero of the polynomial f(x), then the divisor of f(x) will be _________

Detailed Solution for Test: Polynomial Division - Question 5

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.

Test: Polynomial Division - Question 6

What real number that should be added to the polynomial f(x) = 81x^{2 }- 31, so that it is exactly divisible by 9x + 1?

Detailed Solution for Test: Polynomial Division - Question 6

81x^{2 }- 31 is exactly divisible by 9x + 1

Hence, on dividing 81x^{2 }- 31 by 9x + 1

We get, 9x - 1 as quotient and remainder as -30.

So if we add 30 to 81x^{2 }- 31, it will be exactly divisible by 9x + 1.

Test: Polynomial Division - Question 7

When a polynomial f(x) = acx^{3 }+ bcx + d, is divided by g(x), it leaves quotient as cx, and remainder as d. The value of g(x)will be _____

Detailed Solution for Test: Polynomial Division - Question 7

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.

acx^{3} + bcx + d = cx × g(x) + d

acx^{3} + bcx + d – d = cx × g(x)

g(x) = ax^{2 }+ b

Test: Polynomial Division - Question 8

The quotient if the polynomial f(x) = 50x^{2 }- 90x - 25 leaves a remainder of -5, when divided by 5x - 10, will be __________

Detailed Solution for Test: Polynomial Division - Question 8

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.

∴ 50x^{2 }- 90x - 25 = q(x) × 5x - 10 - 5

50x^{2 }- 90x - 25 + 5 = q(x) × 5x - 10

We get, q(x) = 10x + 2

Test: Polynomial Division - Question 9

If two of the zeros of the polynomial f(x) = x^{3 }+ (6 - √3)x^{2 }+ (-1 - √3)x + 30 - 6√3 are 3 and -2 then, the other zero will be ____________

Detailed Solution for Test: Polynomial Division - Question 9

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) = x^{2 }+ 2x - 3x - 6 = x^{2 }- x + 6

Dividing, x^{3 }+ (6 - √3)x^{2 }+ (-1 - √3)x + 30 - 6√3 by x^{2 }- x + 6

We get, x - 5 + √3 as quotient.

Hence, the third zero will be 5 - √3.

Test: Polynomial Division - Question 10

If f(x) is divided by g(x), it gives quotient as q(x) and remainder as r(x). Then, 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.

Detailed Solution for Test: Polynomial Division - Question 10

Consider, f(x) is 27x^{2 }- 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 = 27x^{2 }- 45x + 6x - 10 + 10 = 27x^{2 }- 39x, which is equal to LHS.

Hence proved.

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