anonymous
  • anonymous
can anyone help me with this question Use the fact that f(-2)=0 to factor f(x)=x^3-3x^2-16x-12
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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caozeyuan
  • caozeyuan
well, if f(-2)=0, x+2 has to be a factor right?
anonymous
  • anonymous
oh yeah i never realized that thanks for pointing that out for me
caozeyuan
  • caozeyuan
the other factor has one degree lower than the given cuz x+2 is first order

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caozeyuan
  • caozeyuan
which means the factor is (x+2)(ax^2+bx+c) and a is not 0
caozeyuan
  • caozeyuan
now set two equal to each other, expand LHS and compare coefficients to determine a,b,c if the secondary polynomial has real roots, rewrite the polynomial as (x+x1)(x+x2) where x1and x2 are the roots
anonymous
  • anonymous
im lost now
caozeyuan
  • caozeyuan
Ok lets work this out, you undertand that x+2 is one of the factors right
anonymous
  • anonymous
yes
caozeyuan
  • caozeyuan
and the other part has the form ax^2+bx+c,right?
anonymous
  • anonymous
where did you get the ax and the bx and the c
caozeyuan
  • caozeyuan
Do you mean how did I come up with this form?
anonymous
  • anonymous
yes
caozeyuan
  • caozeyuan
well, 5^3 can be factored into 5^2 and 5, right
caozeyuan
  • caozeyuan
same thing here, we pull a x out of x^3,the left part has to be x^2
anonymous
  • anonymous
ok
caozeyuan
  • caozeyuan
so we have the answer already, it is (x+2)(ax^2+bx+c)
caozeyuan
  • caozeyuan
therefore it has to equal to x^3-3x^2-16x-12
anonymous
  • anonymous
so whats the answer
caozeyuan
  • caozeyuan
we are getting there, still a long way to go actually
caozeyuan
  • caozeyuan
so we have (x+2)(ax^2+bx+c)=x^3-3x^2-16x-12
caozeyuan
  • caozeyuan
therefore ax^3+(2a+b)x^2+(2b+c)x+2c=RHS
caozeyuan
  • caozeyuan
now we have a=1,2c=-12,2a+b=-3, so you can solve for a b and c
anonymous
  • anonymous
ok
caozeyuan
  • caozeyuan
so tell me what are a b and c
anonymous
  • anonymous
b=-5
caozeyuan
  • caozeyuan
ok, good
caozeyuan
  • caozeyuan
now set ax^2+bx+c=0, is there any real root?
anonymous
  • anonymous
no
caozeyuan
  • caozeyuan
well, yes actually b^2-4ac text remember?
anonymous
  • anonymous
yeah sorry
caozeyuan
  • caozeyuan
so (-5)^2-4*-6=49, which is neat, because it is 7^2
caozeyuan
  • caozeyuan
now can you solve the equation?
anonymous
  • anonymous
yes
caozeyuan
  • caozeyuan
so what is the solution?
anonymous
  • anonymous
isn't 49
caozeyuan
  • caozeyuan
I am the quadratic equation what are the roots, there are two, if you haven't realized already
anonymous
  • anonymous
well thanks for the help i got to go now

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