How do you divide $(2 k^{3} - 13 k^{2} - 77 k + 60) \div (k - 10)$ $(2k^3-13k^2-77k+60)div(k-10)$ using synthetic division?

Question

2126 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-06T20:07:38

Quotient is $2 k^{2} + 7 k - 7$ $2k^2+7k-7$ and remainder is $- 10$ $-10$

$2 k^{3} - 13 k^{2} - 77 k + 60$ $2k^3-13k^2-77k+60$

= $2 k^{3} - 20 k^{2} + 7 k^{2} - 70 k - 7 k + 70 - 10$ $2k^3-20k^2+7k^2-70k-7k+70-10$

= $2 k^{2} \cdot (k - 10) + 7 k \cdot (k - 10) - 7 \cdot (k - 10) - 10$ $2k^2*(k-10)+7k*(k-10)-7*(k-10)-10$

= $(2 k^{2} + 7 k - 7) \cdot (k - 10) - 10$ $(2k^2+7k-7)*(k-10)-10$

Hence quotient is $2 k^{2} + 7 k - 7$ $2k^2+7k-7$ and remainder is $- 10$ $-10$

How do you divide (2k^3-13k^2-77k+60)div(k-10)(2k3−13k2−77k+60)÷(k−10)(2k^3-13k^2-77k+60)div(k-10) using synthetic division?