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  • Check whether p(x) is multiple of g(x) or not: (i)p(x)=x^{3}-5x^{2}+4x-3, g(x)=x-2\\ (ii) p(x)=2x^{3}-11x^{2}-4x+5, g(x)=2x+1

Check whether p(x)  is multiple of g(x)  or not

\\(i)p(x)=x^{3}-5x^{2}+4x-3, g(x)=x-2\\ (ii) p(x)=2x^{3}-11x^{2}-4x+5, g(x)=2x+1

Answers (1)

(i) No

Solution

Given p(x)=x^{3}-5x^{2}+4x-3, g(x)=x-2

According to remainder theorem if p(x)  is a multiple of (x+a)  then p(-a)=0

Here p(x) is  x^{3}-5x^{2}+4x-3   and g(x)  is x-2

If p(x)  is a multiple of g(x)  then p(2)=0
\\p(2)=(2)^{3}-5(2)^{2}+4(2)-3\\ =8-20+8-3\\ =16-23=-7\neq 0

p(2)\neq 0

Hence p(x)  is not a multiple of g(x) .

(ii) No

Solution

Given p(x)=2x^{3}-11x^{2}-4x+5, g(x)=2x+1

According to remainder theorem if p(x)  is a multiple of (x+a)  then p(-a)=0

If p(x)  is a multiple of g(x)  then p\left (-\frac{1}{2} \right )=0
\\p\left (-\frac{1}{2} \right )=2\left (-\frac{1}{2} \right )^{3}-11\left (-\frac{1}{2} \right )^{2}-4\left (-\frac{1}{2} \right )+5\\ =2 \times \frac{-1}{8}-11 \times \frac{1}{4}+2+5\\ =-\frac{1}{4}-\frac{11}{4}+7\\ =\frac{-1-11+28}{4}=\frac{16}{4}=4\\ p\left (-\frac{1}{2} \right )\neq 0

Hence p(x)  is not a multiple of g(x) .

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