How do you divide #(x^3-11x^2+22x+40)div(x-5)# using synthetic division?

1 Answer
Sep 18, 2016

Answer:

#x^2-6x-8#

Explanation:

The coefficients of the first polynomial become the dividend.
The "zero" of the second polynomial becomes the divisor. In other words, solve #x-5=0# and use the result (5) as the divisor.

#5__|1color(white)(aa)-11color(white)(aaaa)22color(white)(aaa)40#
#color(white)(aaAaaaaa)5color(white)(a)-30color(white)(a)-40#
#color(white)(aa)#-----------------------------------
#color(white)(aa)color(red)1color(white)(aaa)color(red)(-6)color(white)(aa)color(red)(-8)color(white)(aaaa)color(red)0#

To do synthetic division, "pull down" the first coefficient (#1#) below the line. Multiply the divisor by the first coefficient. #5*1=5#

Write the product (#5#) under the next coefficient and add. (#-11+5=-6#) Write the sum under the line.

Multiply the divisor by the new number under the line.
(#5*-6=-30#)

Write the product (-30) under the next coefficient and add #22+ -30=-8#

Continue as above. The remainder is zero.

Then, use the numbers you've written under the line as the coefficients of the quotient. Start with a variable of degree one less than the degree of the dividend.

The quotient is:
#color(red)1x^2color(red)(-6)xcolor(red)(-8)+ color(red)0/(x^3-11x^2+22x+40)#

The last term represents the remainder and is equal to zero.

Thus, the quotient is:
#x^2-6x-8#