Question 318247
1. {{{(32x^3-48x^2-56x-12)/ (8x+4)}}}
Factor out 4
{{{(4(8x^3-12x^2-14x-3))/ (4(2x+1))}}}
cancel the 4's
{{{(8x^3-12x^2-14x-3)/ (2x+1)}}}
using long division:
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(2x+1)| 8x^3 - 12x^2 - 14x - 3
you will get quotient of: 4x^2 - 8x - 3
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 2.I have to convert this to a decimal notation and am not sure how to go about it 
10^-3 = .001; prove this on your calc; enter 10^-3
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3.r^2+15r = 0 I am looking for the solution to r how would I find it?
Factor out r
r(r + 15) = 0
two solutions
r = 0
r = -15
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 4. factor by grouping gives me trouble could you show me how to do it? 3x^3-24x^2-5x+40
Factor out 3x^2 on the 1st two terms, factor out 5 on the last two terms
Note when you factor out -5, it makes it a MINUS 8
3x^2(x-8) - 5(x-8)
Factor out (x-8) from each term; you are left with:
(x-8)(3x^2 - 5)
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;
 5.(r-45+7)+(5r+3s)+(s+2)
plus in front of the brackets so just remove them:
r-45+7+5r+3s+s+2
Group like terms
r + 5r + 3s + s - 45 + 7 + 2
6r + 4s - 36
Factor out 2
2(3r + 2s - 18)