How do you factor #-5k^{2}-22k+15#?

2 Answers
Jim G.
Apr 23, 2018

#(+5+k)(3-5k)#

Explanation:

#"using the a-c method of factoring"#

#"the factors of "-5xx15=-75"#

#"which sum to - 22 are - 25 and + 3"#

#"split the middle term using these factors"#

#=-5k^2-25k+3k+15larrcolor(blue)"factor by grouping"#

#=color(red)(-5k)(k+5)color(red)(+3)(k+5)#

#"take out the "color(blue)"common factor "(k+5)#

#=(k+5)(color(red)(-5k+3))#

#rArr-5k^2-22k+15=(5+k)(3-5k)#

EZ as pi
Apr 23, 2018

#=-(5k-3)(k+5)#

Or with re-arranging:

#(3-5k)(5+k)#

Explanation:

#-5k^2-22k+15#

The negative at the front is not comfortable foe factorising.
Divide it out as a common factor. All the signs will change.

#=-(5k^2+22k-15)#

The factors of a quadratic trinomial are two brackets.

The negative in front of #15# tells you two things:

  • find factors of #5 and 15# whose products SUBTRACT
    to make #22#
  • the signs in the two brackets will be DIFFERENT

A quick thought is that #5xx5 = 25# which is close to #22#

#" "5 and 15#
#" "darr" "darr#
#" "5" "3" "rarr 1 xx 3 = 3#
#" "1" "5" "rarr 5 xx 5 = ul25#
#color(white)(xxxxxxxxxxxxxxxxx)22" "(larr 25-3=22)#

We have the correct factors, now include the signs to get #-22#

#" "5 and 15#
#" "darr" "darr#
#" "5" "-3" "rarr 1 xx color(red)(-3 = -3)#
#" "1" "+5" "rarr 5 xx color(blue)(+5 = ul(+25))#
#color(white)(xxx.xxxxxxxxxxxxxxxx)color(blue)(+22)" "(larr +25-3=+22)#

The top row gives the first bracket and the bottom row gives the second bracket:

#=-(5k-3)(k+5)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
You could also solve the sign issue by re-arranging the terms:

#15-22k -5k^2#

This leads to the factors:

#(3-5k)(5+k)#

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