<p>I did this problem from a RUSH practice test:</p>
<p>What is the sum of the zeros of f(x) = (x-2)(x^2-4)?</p>
<p>I thought that the roots were: 2,-2, so the sum of those roots would be 0. Rush, on the other hand, counted the 2 twice because it was a root from both the x-2 and x^2-4 equation, so they said the answer was 2+2-2=2.</p>
<p>Who made the mistake, me or RUSH? I heard that there are a couple of errors here and there, and I found this one in the first practice test.</p>
<p>RUSH is right.</p>
<p>When you sum the zeroes, you have to include any duplicate zeroes. It’s a third degree function, so it should have 3 zeroes.</p>
<p>well for (x-2) = 0, the zero is 2…and for (x^2-4), the zeros are 2 and -2 because it’s x squared…therefore it IS 2+2-2. you count all of the zeros</p>
<p>No, you’re right. In this case, x = 2 is a duplicate root. The function only has two zeros; you can’t just repeat the same answer an indefinite amount of times.</p>
<p>^ It has 3 zeroes total. Only 2 distinct zeroes.</p>
<p>A third degree polynomial will have three zeroes, real or complex. You include both 2’s.</p>
<p>(x-2)(x^2-4)
x^3-2x^2-4x+8</p>
<p>handy formulas to know.
sum of roots is -b/a
product of roots is -constant/a</p>
<p>-(-2)/(1)=2</p>
<p>proof: since the roots are -2 and 2(since this root only touches the X axis, it is repeated)
sum would be -2+2+2=2
product would be -2x2x2 which would be -8.</p>
<p>using formulas is alot easier eh?</p>