Question

6 multiple choice 1 point which are the solutions? $2x^3 - 3x = 5x^2$ $0, 1, \frac{3}{2}$ $0, -\frac{1}{2}, 3$ $0, \frac{1}{2}, -3$ $0, -1, -\frac{3}{2}$ $-\frac{1}{2}, 3$ clear my selection 7 fill in the blank 1 point factor completely. $5x^3 - 8x^2 + 15x - 24$ fill in the values of the letters in the answer: $(ax^2 + b)(cx + d)$ a= type your answer... b= type your answer... c= type your answer... d= type your answer...

Explanation:

Question 6

Step1: Rearrange the equation

First, we rearrange the given equation \(2x^{3}-3x = 5x^{2}\) to the standard form of a polynomial equation. We subtract \(5x^{2}\) from both sides to get:
\(2x^{3}-5x^{2}-3x = 0\)

Step2: Factor out the common factor

We can factor out an \(x\) from each term:
\(x(2x^{2}-5x - 3)=0\)

Step3: Factor the quadratic

Now we factor the quadratic \(2x^{2}-5x - 3\). We need two numbers that multiply to \(2\times(-3)=-6\) and add up to \(-5\). The numbers are \(-6\) and \(1\). So we rewrite the middle term:
\(2x^{2}-6x+x - 3 = 0\)
Then we factor by grouping:
\(2x(x - 3)+1(x - 3)=0\)
\((2x + 1)(x - 3)=0\)

Step4: Find the solutions

From the factored form \(x(2x + 1)(x - 3)=0\), we set each factor equal to zero:

• \(x = 0\)
• \(2x+1 = 0\Rightarrow x=-\frac{1}{2}\)
• \(x - 3=0\Rightarrow x = 3\)

Step1: Group the terms

We have the polynomial \(5x^{3}-8x^{2}+15x - 24\). We group the first two terms and the last two terms:
\((5x^{3}-8x^{2})+(15x - 24)\)

Step2: Factor out the common factors from each group

From the first group \(5x^{3}-8x^{2}\), we can factor out \(x^{2}\) (wait, no, let's check again. Wait, actually, if we group as \((5x^{3}+15x)+(-8x^{2}-24)\), that might be better. Let's try that.
Grouping as \((5x^{3}+15x)+(-8x^{2}-24)\)
Factor out \(5x\) from the first group: \(5x(x^{2}+3)\)
Factor out \(-8\) from the second group: \(-8(x^{2}+3)\)
Now we have:
\(5x(x^{2}+3)-8(x^{2}+3)\)
Then we factor out \((x^{2}+3)\):
\((x^{2}+3)(5x - 8)\)? Wait, no, the problem says the form is \((Ax^{2}+B)(Cx + D)\). Wait, maybe my initial grouping was wrong. Let's go back. The original polynomial is \(5x^{3}-8x^{2}+15x - 24\). Let's group as \((5x^{3}-8x^{2})+(15x - 24)\). Factor \(x^{2}\) from the first group: \(x^{2}(5x - 8)\), factor \(3\) from the second group: \(3(5x - 8)\). Ah, there we go. So:
\(x^{2}(5x - 8)+3(5x - 8)=(x^{2}+3)(5x - 8)\)

Wait, but the problem says the form is \((Ax^{2}+B)(Cx + D)\). So comparing \((x^{2}+3)(5x - 8)\) with \((Ax^{2}+B)(Cx + D)\), we have:

• \(A = 1\) (coefficient of \(x^{2}\) in the first factor)
• \(B = 3\) (constant term in the first factor)
• \(C = 5\) (coefficient of \(x\) in the second factor)
• \(D=-8\) (constant term in the second factor)

Wait, but let's check the multiplication: \((x^{2}+3)(5x - 8)=5x^{3}-8x^{2}+15x - 24\), which matches the original polynomial. So:

Step1: Group and factor

Group the polynomial as \((5x^{3}-8x^{2})+(15x - 24)\)
Factor \(x^{2}\) from the first group: \(x^{2}(5x - 8)\)
Factor \(3\) from the second group: \(3(5x - 8)\)
Then factor out \((5x - 8)\): \((x^{2}+3)(5x - 8)\)

Step2: Identify \(A\), \(B\), \(C\), \(D\)

Comparing \((x^{2}+3)(5x - 8)\) with \((Ax^{2}+B)(Cx + D)\):

• \(A = 1\) (since the coefficient of \(x^{2}\) in the first factor is \(1\))
• \(B = 3\) (the constant term in the first factor is \(3\))
• \(C = 5\) (the coefficient of \(x\) in the second factor is \(5\))
• \(D=-8\) (the constant term in the second factor is \(-8\))

Answer:

\(0,-\frac{1}{2},3\) (the option with this set of solutions)

Question 7