How do you find the quotient of #(b^2+6b+5)/(6b+6)div(b+5)#?

Question

Eva Abramson · Accepted Answer

#1/6#

The first step with algebraic fractions is to factorise wherever possible:

#(b^2 +6b+5)/(6b+6) color(blue)(div ((b+5))/1)#

#=((b+5)(b+1))/(6(b+1)) color(blue)(xx 1/((b+5)))" "larr# multiply by the reciprocal

#=(cancel((b+5))cancel((b+1)))/(6cancel((b+1))) xx 1/(cancel((b+5)))" "larr# cancel like factors

#=1/6#

Eva Adamson · Answer

undefined To find the quotient of (b^2+6b+5)/(6b+6) divided by (b+5), you can use polynomial long division. Here are the steps:

1. Divide the first term of the numerator (b^2) by the first term of the denominator (b). The result is b.
2. Multiply the entire denominator (b+5) by the quotient obtained in step 1 (b), and subtract it from the numerator (b^2+6b+5).
   (b^2+6b+5) - (b(b+5)) = b^2+6b+5 - (b^2+5b) = b+5b+5 = 6b+5
3. Repeat steps 1 and 2 with the new numerator (6b+5) and the denominator (b+5).
   Divide the first term of the new numerator (6b) by the first term of the denominator (b). The result is 6.
   Multiply the entire denominator (b+5) by the quotient obtained in step 3 (6), and subtract it from the new numerator (6b+5).
   (6b+5) - (6(b+5)) = 6b+5 - (6b+30) = -25
4. Since the new numerator (-25) is a constant and has a degree lower than the denominator (b+5), the division is complete.
   The quotient is b+6 with a remainder of -25.

Therefore, the quotient of (b^2+6b+5)/(6b+6) divided by (b+5) is b+6 with a remainder of -25.