anonymous
  • anonymous
1. Determine the zeros of f(x) = x3 – 3x2 – 16x + 48
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
soryy it took so long @ganeshie8
ganeshie8
  • ganeshie8
aha its okay :)
ganeshie8
  • ganeshie8
\( f(x) = x^3 – 3x^2 – 16x + 48 \)

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ganeshie8
  • ganeshie8
you have four terms.
ganeshie8
  • ganeshie8
look at first two terms. we can pull out x^2 ?
anonymous
  • anonymous
yes so x²(x-3) ?
ganeshie8
  • ganeshie8
thats right ! next, look at last two terms, can we pull out -16 ?
ganeshie8
  • ganeshie8
\( f(x) = x^3 – 3x^2 – 16x + 48 \) \( = x^2(x-3) - 16(x - 3)\)
anonymous
  • anonymous
okk , I get that part.
ganeshie8
  • ganeshie8
good :) next, see (x-3) is common
ganeshie8
  • ganeshie8
pull it out
ganeshie8
  • ganeshie8
\( f(x) = x^3 – 3x^2 – 16x + 48 \) \( = x^2(x-3) - 16(x - 3)\) \( = (x-3)(x^2 - 16)\)
anonymous
  • anonymous
(x²-16)(x-3)
ganeshie8
  • ganeshie8
Now, to get the zeroes set it to 0
anonymous
  • anonymous
sooo 4 and 3 ?
ganeshie8
  • ganeshie8
\((x-3)(x^2 - 16) = 0 \) \((x-3) = 0 , \ \ (x^2 - 16) = 0 \) \(x = 3 , \ \ x^2 = 16 \) \(x = 3 , \ \ x = \pm\sqrt{16} \) \(x = 3 , \ \ x = 4, \ \ \ x = -4 \)
ganeshie8
  • ganeshie8
3 zeroes, since it is a 3rd degree polynomial
anonymous
  • anonymous
ohhh okk thankyouu veryy much, i understand it now (: . okk im going to try and do the next 3 and ill tag you in them (;
ganeshie8
  • ganeshie8
brilliant :)

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