<li><p>Between which two consecutive integers is there a zero of P(x) = 28x^3 - 11x^2 +15x - 28?
a) -2 and -1
b) -1 and 0
c) 0 and 1
d) 1 and 2
e) 2 and 3</p></li>
<li><p>If i is a root of x^4 + 2x^3 - 3x^2 + 2x - 4 = 0, what are the real roots?
a) ± 2
b) 1, -4
c) -1 ± rad 5
d) 1 ±rad5
e) -1, 4</p></li>
<li><p>If 3x^3 - 9x^2 + Kx -12 is divisible by x-3, then it is also divisible by
a) 3x+2 -x +4
b) 3x^2 -4
c) 3x^2 + 4
d) 3x-4
e) 3x + 4</p></li>
</ol>
<p>4.WRite teh equation of lowest degree with real coefficients if two of its roots are -1 and 1 + i.
a) x^3 + x^2 + 2 = 0
b) x^3 -x^2 -2 =0
c) x^3 - x + 2 = 0
d) x^3 -x^2 +2 = 0
e) none of the above</p>
<p>Answers are C,C,C,D. Please explain how to get these answers</p>
<p>3) Use the factor theorm to figure out K
3(3)^3 - 9(3)^2 + K(3) -12 = 0
3k - 12 = 0
3k = 12
k = 4</p>
<p>Now you can figure out which of answers divides evenly into 3x^3 - 9x^2 + 4x -12. (There's probably a faster, more "SAT friendly" way to figure it out which of the choices is right, but at this point, with my knowledge, I would use trial and error)</p>
<p>1) If you can use a graphing calculator (I don't know what the policies are regarding calculators in the exam, so I apologize if this is irrelevant), you can graph the equation and calculate the "zero" (the x-intercept)</p>
<ol>
<li><p>P(-2), P(-1), P(0) are all < 0, and P(1),P(2),P(3) are all > 0. P(x) changes sign between x=0 and 1, so the answer is C.</p></li>
<li><p>If i is a root, then -i is also a root; so (x^2 - i^2) or (x^2 +1) is a factor. You can divide x^4 + 2x^3 - 3x^2 + 2x - 4 by (x^2+1) to get a quadratic in x, and find the roots.</p></li>
</ol>
<p>Another way is by elimination. Try x=1, then f(x) <> 0, so B is out. For x=-1, f(x) is again <>0, so E is out....</p>
<ol>
<li>If x= 1 + i is a root, then x=1 - i is also a root.
The polynomial is (x+1) (x -1-i) (x -1 + i)
= (x+1) ( x^2 -2x + 2)
= x^3 -x^2 +2</li>
</ol>
<ol>
<li>If i is a root, then -i is also a root; so (x^2 - i^2) or (x^2 +1) is a factor. </li>
</ol>
<p>Why did u use (x^2 - i^2) as a factor?</p>
<p>In general, if x=a is a root for a polynomial, then (x-a) is a factor. In problem(2) above, if x=i and x=-i are roots, then (x-i) and (x+i) are factors, so the polynomial can be written as (x-i)(x+i) (some other function), or
(x^2 - i^2) (some other function).</p>