Q. Multiply and simplify: \( (2x – 3)(3x^2 + x – 4) \)
Answer
Multiply term-by-term:
\[
(2x-3)(3x^2+x-4)=2x(3x^2+x-4)-3(3x^2+x-4)
\]
\[
=6x^3+2x^2-8x-9x^2-3x+12
\]
Combine like terms:
\[
=6x^3-7x^2-11x+12.
\]
Final result: \(6x^3-7x^2-11x+12\).
Detailed Explanation
Expand and simplify the product \( (2x-3)(3x^2+x-4) \).
-
Use the distributive property (multiply each term of the first factor by the entire second factor):
First multiply \(2x\) by each term of \(3x^2+x-4\):
\[
2x\cdot(3x^2+x-4)=2x\cdot3x^2+2x\cdot x+2x\cdot(-4)
=6x^3+2x^2-8x
\] -
Now multiply \(-3\) by each term of \(3x^2+x-4\):
\[
-3\cdot(3x^2+x-4)=-3\cdot3x^2-3\cdot x-3\cdot(-4)
=-9x^2-3x+12
\] -
Add the two results and combine like terms:
\[
(6x^3+2x^2-8x)+(-9x^2-3x+12)
=6x^3+(2x^2-9x^2)+(-8x-3x)+12
=6x^3-7x^2-11x+12
\]
Final answer: \(\;6x^3-7x^2-11x+12\)
FAQs
Q What is the expanded form of \( (2x-3)(3x^2+x-4) \)?
Q Which method is best: FOIL or distributive?
Q Can the cubic be factored further?
Q What are the roots/zeros of the polynomial?
Q What is the degree and leading coefficient?
Q What is the end behavior of the function \(f(x)=6x^3-7x^2-11x+12\)?
Q Common mistakes to avoid when expanding?
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