1.
Factor completely: \(-3x^3-3x^2+6x\)
Step 1. Factor out the greatest common factor. \[\,-3x^3-3x^2+6x=-3x(x^2+x-2)\,\] Step 2. Factor the trinomial \(x^2+x-2\). We need two numbers with product \(-2\) and sum \(1\): \[2+(-1)=1,\quad 2(-1)=-2\] Step 3. Factor the trinomial. \[x^2+x-2=(x+2)(x-1)\] Step 4. Write the complete factorization. \[-3x^3-3x^2+6x=-3x(x+2)(x-1)\] Answer: B
2.
Factor the following completely: \(81x^2-49y^2\).
Step 1. Recognize a difference of squares. \[81x^2-49y^2=(9x)^2-(7y)^2\] Step 2. Use the formula: \[a^2-b^2=(a-b)(a+b)\] Step 3. Substitute \(a=9x\) and \(b=7y\). \[81x^2-49y^2=(9x-7y)(9x+7y)\] Answer: B
3.
Factor the following completely: \(4x^2-14x-30\).
Step 1. Factor out the greatest common factor. \[4x^2-14x-30=2(2x^2-7x-15)\] Step 2. Factor \(2x^2-7x-15\). We need two numbers with product \(2(-15)=-30\) and sum \(-7\). \[-10+3=-7,\quad (-10)(3)=-30\] Step 3. Split the middle term. \[2x^2-7x-15=2x^2-10x+3x-15\] Step 4. Factor by grouping. \[2x(x-5)+3(x-5)=(2x+3)(x-5)\] Step 5. Include the GCF. \[4x^2-14x-30=2(2x+3)(x-5)\] Answer: D
4.
Find the remainder when the polynomial \(P(x)=2x^3-13x^2+11x+26\) is divided by the binomial \(d(x)=x-5\).
Step 1. Use the Remainder Theorem. If \(P(x)\) is divided by \(x-a\), the remainder is \(P(a)\). Step 2. Since \(d(x)=x-5\), we have \(a=5\). Step 3. Evaluate \(P(5)\). \[\begin{aligned}P(5)&=2(5)^3-13(5)^2+11(5)+26\\&=250-325+55+26\\&=6\end{aligned}\] Therefore, the remainder is \(6\). Answer: C
5.
Determine the remainder when \(P(x)=4x^3-6x^2+4x-3\) is divided by \(2x-1\).
Step 1. Set the divisor equal to zero. \[2x-1=0\] Step 2. Solve for \(x\). \[x=\frac{1}{2}\] Step 3. Use the Remainder Theorem and evaluate \(P\left(\frac{1}{2}\right)\). \[\begin{aligned}P\left(\frac{1}{2}\right)&=4\left(\frac{1}{2}\right)^3-6\left(\frac{1}{2}\right)^2+4\left(\frac{1}{2}\right)-3\\&=4\left(\frac{1}{8}\right)-6\left(\frac{1}{4}\right)+2-3\\&=\frac{1}{2}-\frac{3}{2}-1\\&=-2\end{aligned}\] Therefore, the remainder is \(-2\). Answer: B
6.
Solve the quadratic equation for \(x\): \(2x^2-13x+15=0\)
Step 1. Factor the quadratic. \[2x^2-13x+15=0\] We need two numbers with product \(2(15)=30\) and sum \(-13\): \[-10+(-3)=-13,\quad (-10)(-3)=30\] Step 2. Split the middle term. \[2x^2-10x-3x+15=0\] Step 3. Factor by grouping. \[2x(x-5)-3(x-5)=0\] \[(x-5)(2x-3)=0\] Step 4. Solve each factor. \[x-5=0 \Rightarrow x=5\] \[2x-3=0 \Rightarrow x=\frac{3}{2}\] Answer: A
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