Q. \( (-x – 10)(x^2 – 2x + 1) \)

Answer

We have \(x^2-2x+1=(x-1)^2\). So
\[
(-x-10)(x^2-2x+1)=(-x-10)(x-1)^2=-(x+10)(x-1)^2.
\]
Expanding gives
\[
(-x-10)(x^2-2x+1)=-x^3-8x^2+19x-10.
\]

Final answers: factored form \(- (x+10)(x-1)^2\); expanded form \(-x^3-8x^2+19x-10\).

Detailed Explanation

We simplify the expression \((-x-10)(x^2-2x+1)\) step by step.

Recognize a perfect square:

\(x^2-2x+1=(x-1)^2.\)
Use distribution (or FOIL) to multiply term-by-term:

\[
(-x-10)(x^2-2x+1)=(-x)(x^2-2x+1)+(-10)(x^2-2x+1).
\]
Multiply each part:

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

\((-10)(x^2-2x+1)=-10x^2+20x-10.\)
Combine like terms:

\[
-x^3+2x^2-x-10x^2+20x-10=-x^3-8x^2+19x-10.
\]
Optional: write in factored form using step 1:

\[
(-x-10)(x^2-2x+1)=-(x+10)(x-1)^2.
\]

Final simplified form: \(-x^3-8x^2+19x-10\) (equivalently \(-(x+10)(x-1)^2\)).

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FAQs

How do I expand \( (-x-10)(x^2-2x+1) \)?

-A: Use distributive property: \((-x)(x^2-2x+1)-10(x^2-2x+1)=-x^3-8x^2+19x-10.\)

Q: Can \(x^2-2x+1\) be simplified?

-A: Yes, \(x^2-2x+1=(x-1)^2\) (a perfect square).

What is the fully factored form?

-A: \((-x-10)(x-1)^2\) can be written as \(- (x+10)(x-1)^2\).

What are the roots/zeros and their multiplicities?

-A: Roots: \(x=1\) with multiplicity 2, and \(x=-10\) with multiplicity 1.

What is the polynomial degree and leading coefficient?

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

What is the end behavior of the graph?

-A: -As \(x\to\infty\), \(f(x)\to-\infty\); as \(x\to-\infty\), \(f(x)\to\infty\). (Odd degree, negative leading coeff.)

What is the y-intercept and value at \(x=1\)?

-A: y-intercept \(f(0)=-10\). \(f(1)=0\) (double root).

How do I sketch the graph quickly?

-A: Plot roots \(x=-10\) (crosses) and \(x=1\) (touches and turns), use end behavior, and y-intercept \(-10\); connect smoothly consistent with cubic shape.

How to avoid sign errors when expanding?

-A: Distribute each term carefully: multiply \(-x\) by all terms, then \(-10\) by all terms, combine like terms; check by factoring back to \(-(x+10)(x-1)^2\).

Can I use synthetic division here?

-A: Yes: synthetic divide the expanded cubic by \((x-1)\) twice to confirm the double root and get the linear factor \(x+10\).

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