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Chapter 5: Problem 78

Find each product. In each case, neither factor is a monomial. $$\left(x^{2}+3 x+1\right)\left(x^{2}-2 x-1\right)$$

Short Answer

Expert verified
The product of \((x^{2}+3 x+1)\) and \((x^{2}-2 x-1)\) is \[x^{4}+x^{3}-6x^{2}-5x-1\]

Step by step solution

01

Distribute the first term in the first binomial

Multiplying \(x^{2}\) from the bracket \((x^{2}+3 x+1)\) by \((x^{2}-2 x-1)\), we get \(x^{4}-2x^{3}-x^{2}\)
02

Distribute the second term in the first binomial

Multiplying \(3x\) from the bracket \((x^{2}+3 x+1)\) by \((x^{2}-2 x-1)\), we get \(3x^{3}-6x^{2} -3x\)
03

Distribute the last term in the first binomial

Multiplying \(1\) from the bracket \((x^{2}+3 x+1)\) by \((x^{2}-2 x-1)\), we get \(x^{2}-2x-1\)
04

Add all the distributions together

Adding all the outcomes from Step 1 to step 3, we get \[x^{4}-2x^{3}-x^{2}+3x^{3}-6x^{2}-3x+x^{2}-2x-1\]
05

Simplify and Collect Like Terms

Combine the like terms to simplify our expression. The final result is \[x^{4}+x^{3}-6x^{2}-5x-1\]

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