Q. Factor \(x(x+1)(x-4)+4(x+1)\) meaning / means.

Q: What is the fully factored form of x(x+1)(x-4)+4(x+1)?

Factor out x+1: (x+1)[x(x-4)+4]= (x+1)(x^2-4x+4)=(x+1)(x-2)^2.

Q: How do you factor it step-by-step?

Factor the common x+1: x(x+1)(x-4)+4(x+1)=(x+1)(x(x-4)+4). Simplify inside: x^2-4x+4, which is (x-2)^2.

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

Solve (x+1)(x-2)^2=0: x=-1 (simple root) and x=2 (double root, multiplicity 2).

Q: How can I verify the factoring is correct?

Expand (x+1)(x-2)^2: (x-2)^2=x^2-4x+4; multiply by x+1 to recover the original expression.

Q: What is the polynomial’s degree and leading coefficient?

The expression is a cubic: degree 3. The leading term is x^3, so the leading coefficient is 1.

Q: How does the double root at x=2 affect the graph?

How does the double root at x=2 affect the graph?

Q: Could you use substitution to spot the perfect square faster?

Yes: note x^2-4x+4=(x-2)^2. Recognizing that inside the factor (x+1)[\cdots] is a perfect square speeds factoring.

Q: Is the factorization valid over the integers and reals?

Yes. (x+1)(x-2)^2 has integer coefficients and is valid over the integers, reals, and complex numbers.

Answer

Factor the common factor (x+1):
\[
x(x+1)(x-4)+4(x+1)=(x+1)\big(x(x-4)+4\big)=(x+1)(x^2-4x+4)=(x+1)(x-2)^2.
\]

Detailed Explanation

Write the expression to factor: \(x(x+1)(x-4)+4(x+1)\).
Look for a common factor. Both terms contain the factor \(x+1\), so factor it out:

\((x+1)\bigl(x(x-4)+4\bigr)\).
Simplify the expression inside the parentheses. Compute \(x(x-4)+4\):

\(x(x-4)+4 = x^2-4x+4.\)
Recognize the quadratic \(x^2-4x+4\) as a perfect square trinomial. It factors as:

\(x^2-4x+4=(x-2)^2.\)
Substitute this back into the factored form to obtain the complete factorization:

\((x+1)(x-2)^2.\)
Optional verification by expansion:

\((x+1)(x-2)^2=(x+1)(x^2-4x+4)=x^3-3x^2+4,\)

and expanding the original expression gives

\(x(x+1)(x-4)+4(x+1)=x^3-3x^2+4,\)

so the factorization is correct.

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FAQs

What is the fully factored form of \(x(x+1)(x-4)+4(x+1)\)?

Factor out \(x+1\): \((x+1)[x(x-4)+4]= (x+1)(x^2-4x+4)=(x+1)(x-2)^2\).

How do you factor it step-by-step?

Factor the common \(x+1\): \(x(x+1)(x-4)+4(x+1)=(x+1)(x(x-4)+4)\). Simplify inside: \(x^2-4x+4\), which is \((x-2)^2\).

What are the roots/zeros of the polynomial?

Solve \((x+1)(x-2)^2=0\): \(x=-1\) (simple root) and \(x=2\) (double root, multiplicity 2).

How can I verify the factoring is correct?

Expand \((x+1)(x-2)^2\): \((x-2)^2=x^2-4x+4\); multiply by \(x+1\) to recover the original expression.

What is the polynomial’s degree and leading coefficient?

The expression is a cubic: degree \(3\). The leading term is \(x^3\), so the leading coefficient is \(1\).

How does the double root at \(x=2\) affect the graph?

Could you use substitution to spot the perfect square faster?

Yes: note \(x^2-4x+4=(x-2)^2\). Recognizing that inside the factor \((x+1)[\cdots]\) is a perfect square speeds factoring.

Is the factorization valid over the integers and reals?

Yes. \((x+1)(x-2)^2\) has integer coefficients and is valid over the integers, reals, and complex numbers.

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