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How do you solve #p( - 5- p ) = 6#?

2 Answers
Mar 22, 2018

Answer:

#p=-3" or "p=-2#

Explanation:

#"distribute and rearrange into "color(blue)"standard form"#

#rArr-5p-p^2=6#

#rArrp^2+5p+6=0larrcolor(blue)"in standard form"#

#"the factors of + 6 which sum to + 5 are + 3 and + 2"#

#rArr(p+3)(p+2)=0#

#"equate each factor to zero and solve for p"#

#p+3=0rArrp=-3#

#p+2=0rArrp=-2#

Mar 22, 2018

Answer:

#p= -2 and -3#

Explanation:

#p(-5-p)=6#

Using FOIL method :

#-5p-p^2=6#

#p^2+5p+6=0#

Factorization (you can also use quadratic formula and completing square):

#(p+3)(p+2)=0#

#p=-3 and -2#