Question

Use synthetic division to find the quotient and the remainder.

(2b^4-6b^3+3b+16)/(b-2)

Answer 1

Divisor : b-2
Dividend : 2b^4-6b^3+3b+16 or 2b^4-6b^3+0b^2+3b+16 .

The coefficients that are not mentioned , you need to consider them as zero . That is why I included 0b^2 .

Take out the coefficients from the above polynomial ( both divisor and dividend) . Make sure to invert the sign of coefficient present in divisor . Ex : Here -2 is the coefficient in divisor . After inversion it becomes +2 :

Now , perform the below steps sequence wise -

• Write down the first coefficient of dividend as it is . Hence we get 2 as the first coefficient of our quotient .
• The first coefficient of quotient is multiplied with +2 ( coeff. Of divisor ) which results in 4. Add this to second coeff. of dividend i.e 6 . The result is -2 .
• Multiply -2 (second coeff. of quotient) with +2 ( coeff. Of divisor ) to get -4 .
• Add this -4 to 0 to get the result as -4 . Zero mentioned here is the third coeff. of dividend .
• Continue this multiplication and addition till the end .

Hence , Quotient = 2b^3 - 2b^2 -4b - 5 ; Remainder = 6

PS : Please upvote if found it useful . Let me know if you still need explanation with the steps performed .