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one factor of p(x) = 2x^3-9x^2-11x+30 please solve it

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cheemha | Student, Undergraduate | (Level 3) eNoter

Posted July 21, 2010 at 4:53 AM via web

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one factor of p(x) = 2x^3-9x^2-11x+30

please solve it

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kjcdb8er | Teacher | (Level 1) Associate Educator

Posted July 21, 2010 at 5:12 AM (Answer #1)

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2 x^3-9 x^2-11 x+30 = 0

(x-5)(2x^2 + x - 6) = 0

(x-5) (x+2) (2 x-3) = 0

Your factors are:

x = 5, x=-2, x = 3/2

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neela | High School Teacher | (Level 3) Valedictorian

Posted July 21, 2010 at 5:18 AM (Answer #2)

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To solve: p(x) = 2x^3-9x^2-11x+30

Solution.

P(-2) = 2(-2)^3-9(-2)^2-11(-2)+30

= -16-36 -11(-2)+30

= -52+52=0

So , x-(-2) = x+2 is a factor.

Therefore p(x )= p(x) = 2x^3-9x^2-11x+30= (x+2)(ax^2+bx+c). But a and c could easily guessed by equating  x^3 and constant terms: a= 2 and c = 15

Therefore,  p(x) = 2x^3-9x^2-11x+30 = (x+2)(2x^2+bx+15) = 0

Equating the coefficients of x on both sides,

-11 = 15x+2k Or

k = (-11-15)/2 = -13.

So the required factors of p(x) are: (x+2) and 2x^2-13x+15.

But 2x^2-13x+15 = 2x^2 -10x-3x+15

=2x(x-5)-3(x-5)

=(x-5)(2x-3), zeros are x=5 and x=3/2.

Therefore the zeros of p(x) are  -2,  3/2 and 5, or are the solutions of p(x) = 0.

 

 

 

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