How do you solve 2p^ { 3} - 3p ^ { 2} - 98p + 147= 02p33p298p+147=0?

2 Answers
Nov 22, 2017

-7, 7,3/27,7,32

Explanation:

2p^3-3p^2-98p+147=02p33p298p+147=0
=(2p^3-3p^2)+(-98p+147)=(2p33p2)+(98p+147)

factor out p^2p2 from 2p^3-3p^22p33p2
factor out -49 from -98p+14798p+147

=p^2(2p-3)(p^2-49)=p2(2p3)(p249)

factor p^2-49p249

=(p+7)(p-7)(2p-3)=(p+7)(p7)(2p3)
p= -7p=7
p=7p=7
p=3/2p=32

Nov 22, 2017

p=3/2" or "p=+-7p=32 or p=±7

Explanation:

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

2p^3-98p-3p^2+147=02p398p3p2+147=0

rArrcolor(red)(2p)(p^2-49)color(red)(-3)(p^2-49)=02p(p249)3(p249)=0

rArr(p^2-49)(color(red)(2p-3))=0(p249)(2p3)=0

rArr(p-7)(p+7)(2p-3)=0larrcolor(blue)"difference of squares"(p7)(p+7)(2p3)=0difference of squares

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

p-7=0rArrp=7p7=0p=7

p+7=0rArrp=-7p+7=0p=7

2p-3=0rArrp=3/22p3=0p=32