anonymous
  • anonymous
Which of the following is a root of the polynomial shown below? f(x)=x^3-4x^2-11x+30 A. -3 B. 0 C. 1 D. 6
Mathematics
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SOLVED
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chestercat
  • chestercat
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UsukiDoll
  • UsukiDoll
\[f(x) = x^3-4x^2-11x+30\] so given that polynomial, we need to use the rational root test. So using the last term (that's 30) find all factors of 30 plug in x = one of the factors of 30 and see if your result is 0 If it's 0, then we have a root If it's not 0 we don't have root and if all roots don't work, then the rational root test fails
anonymous
  • anonymous
is it C?
UsukiDoll
  • UsukiDoll
oh... we have multiple choice... that's good. We can use either -3,1, or 6. Hint: 0 doesn't work. Is it C? let's try it let x = 1 \[f(1) = 1^3-4(1)^2-11(1)+30\] try computing f(1)

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anonymous
  • anonymous
I got 1-16-11+30 = 0 unless I did something wrong
UsukiDoll
  • UsukiDoll
something isn't right for the 4(1)^2 part
UsukiDoll
  • UsukiDoll
(1)^2 means write 1 twice (1)(1) so for that part 4(1)(1) what's 4 x 1 x 1 ?
anonymous
  • anonymous
oh oops is it 1-4-11+30=16 so it's not c?
UsukiDoll
  • UsukiDoll
not c. so cross that out
UsukiDoll
  • UsukiDoll
so it's either -3 or 6
anonymous
  • anonymous
Oh I got it. I think it's a? -27-36+33-30=-0
anonymous
  • anonymous
oops 0*
UsukiDoll
  • UsukiDoll
\[f(-3) = (-3)^3-4(-3)^2-11(-3)+30\] \[f(-3) = -27-4(9)+33+30\] \[f(-3) = -27-36+33+30\] \[f(-3) = -63+63\] \[f(-3) = 0\] so yes -3 is a root of the polynomial you can just write 0. a negative sign isn't required.
anonymous
  • anonymous
okay thanks!! I get it now :)
UsukiDoll
  • UsukiDoll
yay :D

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