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Chapter 5: Problem 76

Find each product. In each case, neither factor is a monomial. $$\left(x-\frac{1}{3}\right)\left(3 x^{3}-6 x^{2}+5 x-9\right)$$

Short Answer

Expert verified
The product of \( \left(x-\frac{1}{3}\right)\left(3 x^{3}-6 x^{2}+5 x-9\right) \) is \( 3 x^{4} -7 x^{3} +7 x^{2}-9 x+3 \).

Step by step solution

01

Apply Distributive Property

Multiply each term in the binomial each term in the trinomial. This will generate four pairs of terms: \[ x*3 x^{3}, x*-6 x^{2}, x*5 x, x*-9, -\frac{1}{3}*3 x^{3}, -\frac{1}{3}*-6 x^{2}, -\frac{1}{3}*5 x, -\frac{1}{3}*-9 \] Simplifying these expressions, we obtain \[ 3 x^{4}, -6 x^{3}, 5 x^{2}, -9 x, -x^{3}, 2 x^{2}, -\frac{5x}{3}, 3 \]
02

Combine like terms

Now, add or subtract the like terms, which are terms that have the same variable and exponent. This will produce the final expression: \[ 3 x^{4} -7 x^{3} +7 x^{2}-9 x+3 \]

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