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Chapter 0: Problem 29

Factor each trinomial, or state that the trinomial is prime. $$4 x^{2}+16 x+15$$

Short Answer

Expert verified
The factored form of the trinomial \(4x^2 + 16x + 15\) is \( (2x+5) (2x+3) \).

Step by step solution

01

Identify the values of a, b and c

From the provided trinomial, we find that \(a=4\), \(b=16\), and \(c=15\).
02

Calculating the product of a and c

Next, multiply \(a\) and \(c\). In this case, \(4 * 15 = 60\). The two numbers we are looking for must multiply to give 60.
03

Find two numbers that fulfill the requirement

The task is to find two numbers that add up to \(b\), which is 16, and multiply to the product calculated in step 2, which is 60. After some trial and error, we see that these numbers are 6 and 10, since \(6 * 10 = 60\) and \(6 + 10 = 16\).
04

Rewrite the original expression

Now rewrite \(16x\) as \(6x + 10x\) in the original trinomial to get \(4x^2 + 6x + 10x + 15\).
05

Group and factor

Group the terms and factor by grouping to get \(2x(2x + 3) + 5(2x + 3)\).
06

Factored form

Finally, since both terms have the common factor \(2x + 3\), factor out \(2x + 3\) to obtain the final factored form \( (2x+5) (2x+3) \).

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Most popular questions from this chapter

Simplify each expression. Assume that all variables represent positive numbers. $$ \left(\frac{x^{\frac{1}{2}} y^{-\frac{7}{4}}}{y^{-\frac{5}{4}}}\right)^{-4} $$

Exercises \(142-144\) will help you prepare for the material covered in the next section. Simplify and express the answer in descending powers of \(x\) : $$2 x\left(x^{2}+4 x+5\right)+3\left(x^{2}+4 x+5\right)$$

Factor and simplify each algebraic expression. $$(x+5)^{-2}-(x+5)^{-2}$$

Determine whether each statement makes sense or does not make sense, and explain your reasoning. Using my calculator, I determined that \(6^{7}=279,936,\) so 6 must be a seventh root of \(279,936\).

Exercises \(142-144\) will help you prepare for the material covered in the next section. Use the distributive property to multiply: $$2 x^{4}\left(8 x^{4}+3 x\right)$$
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