anonymous
  • anonymous
use the given zero to find the remaining zeros of the function. f(x) = x^3-2x^2-11x+52; zero: -4
Precalculus
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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jdoe0001
  • jdoe0001
so, we know that a zero/solution/root is -4, or that is x = -4 meaning (x+4) x = -4 => (x+4) = 0 if you divide the polynomial by any of its roots, you'd get a zero as remainder so your quotient, will be the one you'd need to factor in order to get the other roots so what would you get for \(\bf \cfrac{ x^3-2x^2-11x+52}{x+4}\)
anonymous
  • anonymous
im not quite sure how i could divide that out
jdoe0001
  • jdoe0001
use long division

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jdoe0001
  • jdoe0001
|dw:1378843583106:dw|
anonymous
  • anonymous
shouldnt be reversed with the polynomial being divided
jdoe0001
  • jdoe0001
well.... yes... the polynomial is the dividend
anonymous
  • anonymous
i get \[x^2-6x+13\] is that close?
jdoe0001
  • jdoe0001
nope, that's correct, thus \(\bf x^3-2x^2-11x+52=0 \implies (x^2-6x+13)(x+4)=0\)
anonymous
  • anonymous
so would i need to do that quadradic formula to find the remaining zeros using the x^2−6x+13
jdoe0001
  • jdoe0001
hmm, yes, you'd have to, because 13 is a prime and won't factor out to get the other 2 roots
anonymous
  • anonymous
thats what i was thinking, thank you for your help :)
jdoe0001
  • jdoe0001
yw
jdoe0001
  • jdoe0001
btw, you'd end up with 2 complex roots
anonymous
  • anonymous
ahhh how do i do that xD
jdoe0001
  • jdoe0001
\(\bf x^3-2x^2-11x+52=0 \implies (x^2-6x+13)(x+4)=0\\ \quad \\ \text{quadratic formula}\\ x= \cfrac{ - b \pm \sqrt { b^2 -4ac}}{2a}\\\quad \\ x^2-6x+13\\ x= \cfrac{ - (-6) \pm \sqrt { (-6)^2 -4(1)(13)}}{2(1)}\)

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