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  • Show that (i) x+3 is a factor of 69+11x-x^{2}+x^{3} (ii) 2x-3 is a factor of x+2x^{3}-9x^{2}+12

Show that

(i)  x+3  is a factor of 69+11x-x^{2}+x^{3}

(ii) 2x-3  is a factor of x+2x^{3}-9x^{2}+12

Answers (1)

(i) Here g(x)=x+3  and p(x)=69+11x-x^{2}+x^{3}

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

So, if g(x) is a factor of p(x) then p(-3)=0
\\p(-3)=69+11(-3)-(-3)^{2}+(-3)^{3}\\ =69-33-9-27\\ =69-69=0\\ p(-3)= 0

Therefore x+3  is a factor of 69+11x-x^{2}+x^{3}

Hence proved

(ii) Here g(x)=2x-3  and p(x)=x+2x^{3}-9x^{2}+12

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

So, if g(x) is a factor of p(x) then p\left ( \frac{3}{2} \right )=0

\\p\left ( \frac{3}{2} \right )=\left ( \frac{3}{2} \right )+2\left ( \frac{3}{2} \right )^{3}-9\left ( \frac{3}{2} \right )^{2}+12\\ =\frac{3}{2}+2 \times \frac{27}{8}-9 \times \frac{9}{4}+12\\ =\frac{3}{2}+\frac{27}{4}-\frac{81}{4}+12\\ =\frac{6+27-81+48}{4}\\ =\frac{0}{4}=0\\ p\left ( \frac{3}{2} \right )=0

Hence 2x-3  is a factor of  x+2x^{3}-9x^{2}+12

Hence proved

 

Posted by

infoexpert24

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