anonymous
  • anonymous
Which is a factor of z3 - z2 - 9z + 9?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
A) (z - 1) B) (z + 1) C) (z - 9) D) (z + 9)
whpalmer4
  • whpalmer4
Well, there's nothing that you can factor out of all of the terms, so it won't be a single variable or a single number. I would guess that you're going to have a \(z+3\) or \(z-3\) or maybe both as factors. \(z+1\) or \(z-1\) might also show up. And \(z+9\) or \(z-9\) might also be possible, although I think not as likely. The reason I can say this is that the presence of the \(z^3\) as the highest power means we'll have 3 things being multiplied together: \[(z+a)(z+b)(z+c)\]or 2 things:\[(z^2 + ...+a)(z+b)\] and those constants at the end of each term all get multiplied together to produce that \(+9\). The only way to do that is with 1,3,9, either all 3 positive, or 2 negative and 1 positive.
anonymous
  • anonymous
z-1 ?

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whpalmer4
  • whpalmer4
Now, you have a list of answer possibilities to choose from. Here's a trick: each true factor is an expression which can be set equal to 0 and solved for the value of the variable. If you substitute that value into the polynomial, the polynomial will evaluate to 0! So, \(z+1\) is one possibility. Let's try it out: \[z+1 = 0\]\[z=-1\]Plug \(z=-1\) into the polynomial and see if you get 0: \[z^3 - z^2 - 9z + 9 = (-1)^3-(-1)^2 - 9(-1) + 9 = -1-1+9+9 = 16\]So \(z+1\) is NOT a factor.
whpalmer4
  • whpalmer4
You suggest \(z-1\) as a possibility, let's check it out: \[z-1 = 0\]\[z = 1\] \[(1)^3 - (1)^2 -9(1) + 9 = 1-1-9+9 = 0\]Bingo!
whpalmer4
  • whpalmer4
At this point, to find the other factors, you could synthetic or long division to divide the polynomial by \((z-1)\) and get a simpler polynomial and repeat the process until you have found all the factors.
whpalmer4
  • whpalmer4
At some point, you will often find that you have a polynomial that you can factor by just looking at it. This one, for example, after you divide out the \((z-1)\) factor is \[z^2-9\]which is a difference of squares and thus can be factored as \[z^2-9 = (z-\sqrt{9})(z+\sqrt{9}) = (z-3)(z+3)\] So \(z=\pm 3\) gives the other two factors. Quickly checking: \[(3)^3 - (3)^2 -9(3) + 9 = 27 - 9 - 27 = 0\checkmark\] \[(-3)^3 - (-3)^2 - 9(-3) + 9 = -27-9 + 27 + 9 = 0\checkmark\]

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