Q. \( (2x-3)(3x^2 – x + 6) \)

Answer

Multiply termwise (distribute):

\[
(2x-3)(3x^2-x+6)=2x(3x^2-x+6)-3(3x^2-x+6).
\]

Compute each part:
\[
=6x^3-2x^2+12x-(9x^2-3x+18).
\]

Combine like terms:
\[
=6x^3-(2x^2+9x^2)+(12x+3x)-18=6x^3-11x^2+15x-18.
\]

Final result: \(\boxed{6x^3-11x^2+15x-18}\).

Detailed Explanation

Write the expression:
\( (2x-3)(3x^2 – x + 6) \)
Use the distributive property (multiply each term of the first factor by the entire second factor):
\(2x(3x^2 – x + 6) – 3(3x^2 – x + 6)\)
Multiply term-by-term:
- \(2x\cdot 3x^2 = 6x^3\)
- \(2x\cdot(-x) = -2x^2\)
- \(2x\cdot 6 = 12x\)
- \(-3\cdot 3x^2 = -9x^2\)
- \(-3\cdot(-x) = 3x\)
- \(-3\cdot 6 = -18\)
Add the results and combine like terms:
\(6x^3 + (-2x^2) + 12x + (-9x^2) + 3x – 18\)

Combine the \(x^2\) terms and the \(x\) terms:

\(6x^3 + (-2x^2 – 9x^2) + (12x + 3x) – 18 = 6x^3 – 11x^2 + 15x – 18\)
Final simplified result:
\(\boxed{6x^3 – 11x^2 + 15x – 18}\)

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Step Help

FAQs

Q1: How do you expand \( (2x-3)(3x^2-x+6) \)?

A1: Multiply termwise: \(2x(3x^2-x+6)-3(3x^2-x+6)=6x^3-2x^2+12x-9x^2+3x-18\). Combine like terms: \(6x^3-11x^2+15x-18\).

Q2: What is the degree and leading coefficient of \(6x^3-11x^2+15x-18\)?

A2: Degree is \(3\). Leading coefficient is \(6\).

Q3: How do I factor the polynomial back?

A3: The given factorization is already factored: \((2x-3)(3x^2-x+6)\). No further nontrivial factorization over the rationals exists because the quadratic has negative discriminant.

Q4: What are the roots/zeros of the polynomial?

A4: From \(2x-3=0\) we get \(x=\tfrac{3}{2}\). For \(3x^2-x+6=0\), discriminant \(1-72=-71\), so complex roots \(x=\dfrac{1\pm i\sqrt{71}}{6}\).

Q5: How can I multiply efficiently (method)?

A5: Use distributive property: multiply \(2x\) by each term of the quadratic, then \(-3\) by each term, then add results. It’s FOIL extended to three terms: first do \(2x\cdot(3x^2-x+6)\) and \(-3\cdot(3x^2-x+6)\).

Q6: Is this a special product (perfect square/sum-difference)?

A6: No. It's not a perfect square, sum/difference of cubes, or other common identity; it's a general cubic obtained by multiplying a linear and an irreducible quadratic.

Q7: How can I check my expansion is correct?

A7: Substitute any convenient \(x\) (e.g., \(x=1\)): original \((2-3)(3-1+6)=-8\); expanded \(6-11+15-18=-8\). Matching values confirm the expansion.

Q8: What is the end behavior/graph shape of the polynomial?

A8: Being a cubic with positive leading coefficient \(6\), as \(x\to\infty\), \(y\to\infty\); as \(x\to-\infty\), \(y\to-\infty\). With one real root and two complex roots, the curve crosses the x-axis exactly once.

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