Question

Solve the ques in the attachment.
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Answer 1

HEYA ,

As u told so m here to answer this :))


Refer answer below

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According to the division algorithm, if p(x) and g(x) are two polynomials with

g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that

p(x) = g(x) × q(x) + r(x),

where r(x) = 0 or degree of r(x) < degree of g(x)

Degree of a polynomial is the highest power of the variable in the polynomial.

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(i) deg p(x) = deg q(x)

Degree of quotient will be equal to degree of dividend when divisor is constant ( i.e., when any polynomial is divided by a constant).

Let us assume the division of 6x^2+x+2 by 2.

Here, p(x) = 6x^2+2x+2

g(x) = 2

q(x) = 3x^2+x+1  and r(x) = 0

Degree of p(x) and q(x) is the same i.e., 2.

Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)

$6 {x}^{2} + 2x + 2 = 2(3 {x}^{2} + x + 1) \\ \\ = > 6 {x}^{2} + 2x + 2$


Thus, the division algorithm is satisfied.

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(ii) deg q(x) = deg r(x)

Let us assume the division of x3 + x by x2,

Here, p(x) = x3 + x

g(x) = x2

q(x) = x and r(x) = x

Clearly, the degree of q(x) and r(x) is the same i.e., 1.

Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)

x3 + x = (x2 ) × x + x

x3 + x = x3 + x

Thus, the division algorithm is satisfied.

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(iii)deg r(x) = 0

Degree of remainder will be 0 when remainder comes to a constant.

Let us assume the division of x3 + 1by x2.

Here, p(x) = x3 + 1

g(x) = x2

q(x) = x and r(x) = 1

Clearly, the degree of r(x) is 0.

Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)

x3 + 1 = (x2 ) × x + 1

x3 + 1 = x3 + 1

Thus, the division algorithm is satisfied.

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Hope it helps u !!!

glad to help u :))

# Nikky


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