Q. \((x-4)(2x^2+5x-3)\)

Q: How do I expand (x-4)(2x^2+5x-3) ?

Multiply and combine like terms: x(2x^2+5x-3)-4(2x^2+5x-3)=2x^3-3x^2-23x+12.

Q: How do I factor 2x^2+5x-3 and factor the whole expression completely?

Factor the quadratic: 2x^2+5x-3=(2x-1)(x+3). So the full factorization is (x-4)(2x-1)(x+3).

Q: What are the zeros / x-intercepts of the polynomial?

Set each factor to zero: x=4, x=\frac{1}{2}, x=-3.

Q: What is the y-intercept?

Evaluate at x=0: y=12. So the y-intercept is (0,12).

Q: Can I use synthetic division with x-4?

Yes. Synthetic division by 4 on 2x^3-3x^2-23x+12 gives quotient 2x^2+5x-3 and remainder 0, confirming x-4 is a factor.

Q: What are the multiplicities of the roots?

Each root 4, \frac{1}{2}, -3 has multiplicity 1, so the graph crosses the x-axis at each zero.

Answer

Use distributive law (multiply each term of the quadratic by x and by −4):

\[
(x-4)(2x^2+5x-3)=x(2x^2+5x-3)-4(2x^2+5x-3)
\]
\[
=2x^3+5x^2-3x-8x^2-20x+12
\]
\[
=2x^3-3x^2-23x+12.
\]

Final result: \(\;2x^3-3x^2-23x+12.\)

Detailed Explanation

Problem: Multiply and simplify \( (x-4)(2x^2+5x-3) \).

Distribute \(x\) across the trinomial:
\(x\cdot(2x^2+5x-3)=2x^3+5x^2-3x\).
Distribute \(-4\) across the trinomial:
\(-4\cdot(2x^2+5x-3)=-8x^2-20x+12\).
Add the results and combine like terms:
\[
(2x^3+5x^2-3x)+(-8x^2-20x+12)
=2x^3+(5x^2-8x^2)+(-3x-20x)+12
=2x^3-3x^2-23x+12.
\]

Final answer: \(\displaystyle 2x^3-3x^2-23x+12\)

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FAQs

How do I expand \( (x-4)(2x^2+5x-3) \)?

Multiply and combine like terms: \(x(2x^2+5x-3)-4(2x^2+5x-3)=2x^3-3x^2-23x+12\).

How do I factor \(2x^2+5x-3\) and factor the whole expression completely?

Factor the quadratic: \(2x^2+5x-3=(2x-1)(x+3)\). So the full factorization is \((x-4)(2x-1)(x+3)\).

What are the zeros / x-intercepts of the polynomial?

Set each factor to zero: \(x=4, x=\frac{1}{2}, x=-3\).

What is the degree and leading coefficient of the expanded polynomial?

Degree is 3 (cubic). Leading coefficient is 2, so the polynomial is \(2x^3-3x^2-23x+12\).

What is the y-intercept?

Evaluate at \(x=0\): \(y=12\). So the y-intercept is \((0,12)\).

What is the end behavior of the graph?

Can I use synthetic division with \(x-4\)?

Yes. Synthetic division by 4 on \(2x^3-3x^2-23x+12\) gives quotient \(2x^2+5x-3\) and remainder 0, confirming \(x-4\) is a factor.

What are the multiplicities of the roots?

Each root \(4, \frac{1}{2}, -3\) has multiplicity 1, so the graph crosses the x-axis at each zero.

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