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

Answer

Use distributivity:

\[
(x+3)(x^2+3x+5)=x(x^2+3x+5)+3(x^2+3x+5)
\]
\[
= x^3+3x^2+5x+3x^2+9x+15
\]
\[
= \boxed{x^3+6x^2+14x+15}
\]

Detailed Explanation

Write the product to expand: \( (x+3)(x^2+3x+5) \).
Distribute the first term \(x\) across the second polynomial:
\(x\cdot x^2 = x^3\),\quad \(x\cdot 3x = 3x^2\),\quad \(x\cdot 5 = 5x\).

So from \(x\) we get \(x^3 + 3x^2 + 5x\).
Distribute the second term \(3\) across the second polynomial:
\(3\cdot x^2 = 3x^2\),\quad \(3\cdot 3x = 9x\),\quad \(3\cdot 5 = 15\).

So from \(3\) we get \(3x^2 + 9x + 15\).
Add the two results term-by-term:
\(x^3 + 3x^2 + 5x + 3x^2 + 9x + 15\).
Combine like terms:
\(x^3 + (3x^2+3x^2) + (5x+9x) + 15 = x^3 + 6x^2 + 14x + 15\).
Final expanded form:
\[ (x+3)(x^2+3x+5) = x^3 + 6x^2 + 14x + 15 \]

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FAQs

How do I expand /(x+3)(x^2+3x+5)/?

-A: Multiply distributively: /x(x^2+3x+5)+3(x^2+3x+5)=x^3+3x^2+5x+3x^2+9x+15/. Combine like terms: /x^3+6x^2+14x+15/.

Can it be factored further?

-A: It already is: /(x+3)(x^2+3x+5)/. The quadratic /x^2+3x+5/ has negative discriminant (/9-20=-11/), so it’s irreducible over the reals.

What are the roots/zeros?

-A: From the factors: /x=-3/. The quadratic gives complex roots /x = (-3 \pm i\sqrt{11})/2/.

What is the polynomial’s degree and leading coefficient?

-A: Degree is 3 (cubic). Leading coefficient is 1 (from /x^3/).

What is the constant term and y-intercept?

-A: Constant term is /15/. The y-intercept is the point /(0,15)/.

What is the end behavior?

-A: -As /x \to \infty/, /f(x)\to \infty/; as /x \to -\infty/, /f(x)\to -\infty/ (odd-degree, positive leading coefficient).

Is the function monotonic or does it have local extrema?

-A: Derivative /f'(x)=3x^2+12x+14/. Its discriminant is negative, so /f'(x)>0/ for all real /x/. The function is strictly increasing with no real local extrema.

How to expand using FOIL-like steps for a binomial times trinomial?

-A: Distribute each term of the binomial: multiply /x/ by every term of the trinomial, then multiply /3/ by every term, and add results. That is the general distributive approach for binomial × trinomial.

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