Question
p(x)=x3+8x2−7x+12 and g(x)=x−1. If p(x) is divided by g(x), it gives q(x) and r(x) as quotient and remainder respectively. If a is the degree of q(x) and b is the degree of r(x), (a−b)=?.
Answer: Option B
:
B
Remainder =r(x)=p(1)=13+8×12−7×1+12=14
The degree of a polynomial is the highest degree of its terms when the polynomial is expressed in its canonical form consisting of a linear combination of monomials.
So, Degree of r(x)=b=0
q(x)=p(x)g(x)
Now , p(x)=x3+8x2−7x+12
g(x)=(x−1)
Performing long division
x2+9x+2x−1)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x3+8x2−7x+12 x3−x2 (−) (+) ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 9x2−7x 9x2−9x (−) (+) ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ +2x+12 2x−2 (−) (+) ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 14
∴q(x)=p(x)g(x)=x2+9x+2
So, degree of q(x)=a=2
Hence, a−b=2−0=2
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:
B
Remainder =r(x)=p(1)=13+8×12−7×1+12=14
The degree of a polynomial is the highest degree of its terms when the polynomial is expressed in its canonical form consisting of a linear combination of monomials.
So, Degree of r(x)=b=0
q(x)=p(x)g(x)
Now , p(x)=x3+8x2−7x+12
g(x)=(x−1)
Performing long division
x2+9x+2x−1)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x3+8x2−7x+12 x3−x2 (−) (+) ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 9x2−7x 9x2−9x (−) (+) ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ +2x+12 2x−2 (−) (+) ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 14
∴q(x)=p(x)g(x)=x2+9x+2
So, degree of q(x)=a=2
Hence, a−b=2−0=2
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