Q. Factor \(x^3 + x^2 + 2x + 2\) by grouping.

Q: How do you factor x^3+x^2+2x+2 by grouping?

Group as x^2(x+1)+2(x+1). Factor the common binomial to get (x+1)(x^2+2).

Q: Why choose those particular groups?

You group terms to produce a common factor. Here both x^2(x+1) and 2(x+1) share (x+1), enabling factoring by grouping.

Q: Are there other methods to factor this polynomial?

Yes: use the Rational Root Theorem (test x=\pm1,\pm2), synthetic division if a root is found, or recognize patterns. All yield (x+1)(x^2+2).

Q: Can x^2+2 be factored further over the reals or rationals?

No. x^2+2 has discriminant -8, so it has no real or rational linear factors and is irreducible over both sets.

Q: What are the roots of x^3+x^2+2x+2?

The roots are x=-1 and x=\pm i\sqrt{2}.

Q: How would the factorization change over the complex numbers?

Over \mathbb{C} it factors into linear factors: (x+1)(x-i\sqrt{2})(x+i\sqrt{2}).

Answer

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

Detailed Explanation

Write down the polynomial you want to factor by grouping: \(x^3 + x^2 + 2x + 2\).
Choose a grouping that pairs terms so a common factor appears in each group. Split the polynomial into two groups: \( (x^3 + x^2) + (2x + 2) \).
Factor the greatest common factor from the first group. For \(x^3 + x^2\), both terms contain \(x^2\), so factor \(x^2\) out: \(x^3 + x^2 = x^2(x + 1)\).
Factor the greatest common factor from the second group. For \(2x + 2\), both terms contain \(2\), so factor \(2\) out: \(2x + 2 = 2(x + 1)\).
Now substitute the factored groups back into the expression: \(x^2(x + 1) + 2(x + 1)\). Observe that both terms share the common binomial factor \(x + 1\).
Factor the common binomial \(x + 1\) out of the sum: \(x^2(x + 1) + 2(x + 1) = (x + 1)(x^2 + 2)\).
Verify by expanding the result to confirm it equals the original polynomial: \((x + 1)(x^2 + 2) = x(x^2 + 2) + 1(x^2 + 2) = x^3 + 2x + x^2 + 2 = x^3 + x^2 + 2x + 2\).
Therefore the factorization by grouping is: \((x + 1)(x^2 + 2)\).

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FAQs

How do you factor \(x^3+x^2+2x+2\) by grouping?

Group as \(x^2(x+1)+2(x+1)\). Factor the common binomial to get \((x+1)(x^2+2)\).

Why choose those particular groups?

You group terms to produce a common factor. Here both \(x^2(x+1)\) and \(2(x+1)\) share \((x+1)\), enabling factoring by grouping.

Are there other methods to factor this polynomial?

Yes: use the Rational Root Theorem (test \(x=\pm1,\pm2\)), synthetic division if a root is found, or recognize patterns. All yield \((x+1)(x^2+2)\).

Can \(x^2+2\) be factored further over the reals or rationals?

No. \(x^2+2\) has discriminant \(-8\), so it has no real or rational linear factors and is irreducible over both sets.

What are the roots of \(x^3+x^2+2x+2\)?

The roots are \(x=-1\) and \(x=\pm i\sqrt{2}\).

How do I check my factorization is correct?

When does factoring by grouping fail, and what then?

Grouping fails if you cannot produce a common binomial after grouping. Then try rearranging terms, splitting a middle term, Rational Root Theorem, or polynomial division.

How would the factorization change over the complex numbers?

Over \(\mathbb{C}\) it factors into linear factors: \((x+1)(x-i\sqrt{2})(x+i\sqrt{2})\).

Factor by grouping: x^3+x^2+2x+2 now.
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