-26x+138-35 are 23, find other zeroes.olynomial athe polynomial616-25x

1.	-26x+138-35 are 23, find other zeroes.olynomial athe polynomial616-25x+10 is divided by another polynomial -2k+a, find kand athe remainder comes out to be2.5 Summary
Answer» Fromx^2−2x+k=0,x^2=2x−k After replacingx^2into2x−kin polynomial and equating remainder intox+a, (x2)^2−6x(2x−k)+16(2x−k)−25x+10=x+a (2x−k)^2−12x^2+6kx+32x−16k−25x+10=x+a (2x−k)^2−12x^2+6kx+7x−16k+10=x+a 4x^2−4xk+k^2−12x^2+(6k+7)x−16k+10=x+a −8x^2−4xk+k^2+(6k+7)x−16k+10=x+a −8(2x−k)−4xk+k^2+(6k+7)x−16k+10=x+a −16x+8k+(2k+7)x+k^2−16k+10=x+a (2k−9)x+k^2−8k+10=x+a After equating coefficients, 2k−9=1anda=k^2−8k+10 Hencek=5anda=−5

-26x+138-35 are 23, find other zeroes.olynomial athe polynomial616-25x+10 is divided by another polynomial -2k+a, find kand athe remainder comes out to be2.5 Summary

Answer»

Fromx^2−2x+k=0,x^2=2x−k

After replacingx^2into2x−kin polynomial and equating remainder intox+a,

(x2)^2−6x(2x−k)+16(2x−k)−25x+10=x+a

(2x−k)^2−12x^2+6kx+32x−16k−25x+10=x+a

(2x−k)^2−12x^2+6kx+7x−16k+10=x+a

4x^2−4xk+k^2−12x^2+(6k+7)x−16k+10=x+a

−8x^2−4xk+k^2+(6k+7)x−16k+10=x+a

−8(2x−k)−4xk+k^2+(6k+7)x−16k+10=x+a

−16x+8k+(2k+7)x+k^2−16k+10=x+a

(2k−9)x+k^2−8k+10=x+a

After equating coefficients,

2k−9=1anda=k^2−8k+10

Hencek=5anda=−5