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Chapter 5: Problem 71

Find each product. In each case, neither factor is a monomial. $$(2 a-3)\left(a^{2}-3 a+5\right)$$

Short Answer

Expert verified
The result of the multiplication is \(2a^3 - 9a^2 + 9a - 15\).

Step by step solution

01

Apply the Distributive Law

First, distribute each term of the first polynomial \((2a - 3)\) into each term of the second polynomial \((a^2 - 3a + 5)\), you get four terms as a result: \(2a * a^2 = 2a^3\)\(2a * -3a = -6a^2\)\-3 * a^2 = -3a^2\) \-3 * -3a = 9a\)So far, we have \(2a^3 - 6a^2 - 3a^2 + 9a\).
02

Combine Like Terms

Combine the similar terms. In this case, that's the two terms that involve \(a^2\):\(2a^3 - 6a^2 - 3a^2 = 2a^3 - 9a^2\)Now, the expression is \(2a^3 - 9a^2 + 9a\).
03

Add Remaining Term

Multiplying \(-3\) in the first polynomial and \(5\) in the second polynomial, we get \(-3*5 = -15\). Add this term to the expression. The entire multiplied expression is then:\(2a^3 - 9a^2 + 9a - 15\)

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Most popular questions from this chapter

Exercises \(110-112\) will help you prepare for the material covered in the next section. In each exercise, perform the long division without using a calculator, and then state the quotient and the remainder. $$. 9 8 \longdiv { 2 5 . 1 8 7 }$$

Perform the indicated operations. $$(y+1)\left(y^{2}-y+1\right)-(y-1)\left(y^{2}+y+1\right)$$

Use a vertical format to find each product. $$\begin{array}{r}n^{4}-n^{3}+n^{2}-n+1 \\\2 n+3 \\\\\hline\end{array}$$

Perform the indicated operations. $$(y+5)^{2}-(y-4)^{2}$$

In each exercise, find the product. $$9 x y\left(3 x y^{2}-y+9\right)$$
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