1.
Factor completely the following expression: \(P(x)=x^4-3x^3-39x^2+83x-42\).
Using rational zeros, \(x=1\) is a root twice. After factoring out \((x-1)^2\), the remaining quadratic is \(x^2-x-42=(x-7)(x+6)\). Therefore, \(P(x)=(x-7)(x+6)(x-1)^2\). Answer: B
2.
Solve for the exact value(s) of \(x\): \((x+2)^2-9=-6\).
Add \(9\) to both sides: \((x+2)^2=3\). Take the square root: \(x+2=\pm\sqrt{3}\). Therefore, \(x=-2\pm\sqrt{3}\). Answer: B
3.
Determine the \(y\)-intercept for the given polynomial, \(y=a(x-1)(x+2)(2x+3)\).
To find the \(y\)-intercept, substitute \(x=0\): \(y=a(0-1)(0+2)(2(0)+3)=a(-1)(2)(3)=-6a\). Answer: C
4.
Write a polynomial equation with roots of \(-1\), \(1\), and \(2\).
The roots \(-1,1,2\) give factors \((x+1)(x-1)(x-2)\). Expanding: \((x^2-1)(x-2)=x^3-2x^2-x+2\). Answer: C
5.
Solve for \(x\): \(P(x)=x^3+6x^2+3x-10=0\).
Test \(x=1\): \(1+6+3-10=0\), so \((x-1)\) is a factor. Dividing gives \(x^2+7x+10=(x+5)(x+2)\). Thus the roots are \(x=-5,-2,1\). Answer: C
6.
Which graph best represents \(y=-x(x+3)^2(x-3)^3\)?
The polynomial has degree \(1+2+3=6\), so both ends point in the same direction. The leading coefficient is negative, so both ends point down. It crosses at \(x=0\), bounces at \(x=-3\) because the multiplicity is \(2\), and crosses/flattens at \(x=3\) because the multiplicity is \(3\). Answer: D
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