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Mathematics 7 Online
OpenStudy (anonymous):

Multiply: 2r^2+r-15 r^2+9r+8 ---------- X ----------- r^2+11r+24 2r^2-3r-5

OpenStudy (anonymous):

Factor all numerators and denominators first then cancel

OpenStudy (anonymous):

for starters: \[(r^2 +11r +24) \]is the same as\[(r+8)(r+3)\] and \[(r^2 +9r +8) \] is the same as \[(r+8)(r+1) \]

OpenStudy (anonymous):

\[2r^2+r-15 \]can is also written as \[(2r-5)(r+3)\] and \[2r^2-3r-5\] can be written as \[(r+1)(2r-5)\] through factoring

OpenStudy (anonymous):

2r^2+r-15 r^2+9r+8 ---------- X ----------- r^2+11r+24 2r^2-3r-5 =(2r-5)(r+3) x (r+1)(r+8) ----------- --------- (r+3)(r+8) (2r+5)(r-1) Any equal factors in the top line as the bottom line can be removed =(2r-5) x (r+1) ----------- --------- 1 (2r+5)(r-1) = (2r-5)(r+1) ----------- (2r+5)(r-1) = (2r^2-3r-5) ----------- (2r^2+3r -5)

OpenStudy (anonymous):

the answer is 1

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