How do you use synthetic division to divide #y^2 + 25# by #y+5#?

1 Answer
Sep 2, 2015

Answer:

Write #y^2+25# as #1, 0, 25# (not forgetting the #0#) and #y+5# as #1, 5#, then proceed in a similar fashion to long division to find quotient #1, -5# meaning #y-5# and remainder #50#

Explanation:

#y^2+25 = y^2+0y+25# is represented by the sequence #1#, #0#, #25#.

#y+5# is represented by the sequence #1#, #5#.

Write out like long division of integers and proceed similarly:
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Write #1#, #0#, #25# under the bar as the dividend and #1#, #5# to the left as the divisor.

Identify #1# as the first term of the quotient - writing it above the bar - choosing it to cause the leading terms to match when the divisor is multiplied by it.

Write out #1 xx (1, 5)# under the dividend and subtract it to get the first term #-5# of a remainder.

Bring down the next term #25# of the dividend alongside it.

Identify #-5# as the second term of the quotient - writing above the bar - choosing it to cause the leading terms to match when the divisor is multiplied by it.

Write out #-5 xx (1, 5)# under the remainder and subtract it to get the remainder #50#.

We stop with this remainder since there are not enough terms remaining to make a sequence with enough terms to be divisible by the divisor.

So we have found:

#(y^2 + 25)/(y+5) = (y-5) + 50/(y+5)#

or if you prefer:

#y^2 + 25 = (y+5)(y-5) + 50#