Set up the synthetic division problem with the coefficients of the polynomial as the dividend and the #color(red)1# as the divisor.
#color(red)1|color(white)(aa)1color(white)(aaa)-8color(white)(aaaa)5color(white)(aa)-7#
#color(white)(aaaa)darr#
#color(white)(aaa^2a)color(blue)1color(white)(aaaaaaaaaaaaaaaaaaa)#Pull down the #color(blue)1#
#color(red)1|color(white)(aa)1color(white)(aaa)-8color(white)(aaaa)5color(white)(aa)-7#
#color(white)(aaaa)darrcolor(white)(aaaaa)color(limegreen)1color(white)(aaaaaaaaaaaaa)#Multiply #color(red)1 *color(blue)1# and write the product
#color(white)(aaa^2a)color(blue)1color(white)(aa^(2)a)color(blue)(-7)color(white)(aaaaaaaaaaaaacolor(limegreen)1# under the 8. Add #-8+color(limegreen)1=color(blue)(-7)#
#color(red)1|color(white)(aa)1color(white)(aaa)-8color(white)(aaaa)5color(white)(aa)-7#
#color(white)(aaaa)darrcolor(white)(aaaaa)color(limegreen)1color(white)(a^(22))color(limegreen)(-7)color(white)(aaaaAa)#Multiply #color(red)1*color(blue)(-7)# and put the product
#color(white)(aaa^2a)color(blue)1color(white)(aaaa)color(blue)(-7)color(white)(a^2a)color(blue)(-2)color(white)(aaaaaa)color(limegreen)(-7)# under the #5#. Add #5+color(limegreen)(-7)=color(blue)(-2)#
#color(red)1|color(white)(aa)1color(white)(aaa)-8color(white)(aaaa)5color(white)(aa)-7#
#color(white)(aaaa)darrcolor(white)(aaaaa)color(limegreen)1color(white)(aa)color(limegreen)(-7)color(white)(aaa)color(limegreen)(-2)color(white)(aa)#Repeat multiplying and adding
#color(white)(aaa^2a)color(blue)1color(white)(aa^(2)a)color(blue)(-7)color(white)(a^11)color(blue)(-2)color(white)(aaa)color(magenta)(-9)#
Note that the last number or remainder is #color(magenta)(-9)#.
According to the remainder theorem, #P(1)=color(magenta)(-9)#
