If the quotient on dividing 2x4+x3−14x2−19x+6 by 2x+1 is x3+ax2−bx−6.
Find the values of a and b, also the remainder
Let p(x)=2x4+x3−14x2−19x+6.
Given that the divisor is 2x+1.
Now we get the value of x from
2x+1=0. Then x=−12
∴ The zero of the divisor is x=−12.
So, 2x4+x3−14x2−19x+6=(x+12){2x3−14x−12}+12
=(2x+1)12(2x3−14x−12)+12
Thus, the quotient is 12(2x2−14x−12)=x3−7x−6 and the remainder is 12.
But, given quotient is x3+ax2−bx−6. Comparing this with the quotient obtained we get, a = 0 and b = 7.
Thus, a=0, b=7, and the remainder is 12.