How do you factor #21r ^ { 2} + 132r + 135#?

2 Answers
smendyka
Aug 13, 2017

See a solution process below:

Explanation:

First, factor a #3# out of each term:

#(3 * 7)r^2 + (3 * 44)r + (3 * 45) =>#

#3(7r^2 + 44r + 45)#

Now, playing with factors of #7# and #45# allows us to factor the term within the parenthesis as:

#3(7r + 9)(r + 5)#

Jim G.
Aug 13, 2017

#3(7r+9)(r+5)#

Explanation:

#"factor out the "color(blue)"common factor"# of 3

#rArr21r^2+132r+135#

#=3(7r^2+44r+45)#

#"'splitting' the middle term and factorising by grouping"#

#7r^2+9r+35r+45larr" 9 r + 35 r = 44 r"#

#=color(red)(r)(7r+9)color(red)(+5)(7r+9)#

#=(7r+9)(color(red)(r+5))#

#rArr21r^2+132r+135=3(7r+9)(r+5)#

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