How do you solve #2p^ { 3} - 3p ^ { 2} - 98p + 147= 0#?

2 Answers
Kiana S
Nov 22, 2017

#-7, 7,3/2#

Explanation:

#2p^3-3p^2-98p+147=0#
#=(2p^3-3p^2)+(-98p+147)#

factor out #p^2# from #2p^3-3p^2#
factor out -49 from #-98p+147#

#=p^2(2p-3)(p^2-49)#

factor #p^2-49#

#=(p+7)(p-7)(2p-3)#
#p= -7#
#p=7#
#p=3/2#

Jim G.
Nov 22, 2017

#p=3/2" or "p=+-7#

Explanation:

#"factorise by "color(blue)"grouping"#

#2p^3-98p-3p^2+147=0#

#rArrcolor(red)(2p)(p^2-49)color(red)(-3)(p^2-49)=0#

#rArr(p^2-49)(color(red)(2p-3))=0#

#rArr(p-7)(p+7)(2p-3)=0larrcolor(blue)"difference of squares"#

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

#p-7=0rArrp=7#

#p+7=0rArrp=-7#

#2p-3=0rArrp=3/2#

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