Chapter SH, Problem 13FOWP

Solution

To factors the polynomial $2x^3+4x^2-8x$ .

  $2x\left(x^2+2x-4\right)$

Given:

The polynomial, $2x^3+4x^2-8x$

Concept Used:

Factoring Trinomials in the form  $x^2+ bx + c$

 To factor a trinomial in the form $x^2+ bx + c$ , find two integers,  $\text{'}r \text{'}\text{and} \text{'}s\text{'}$ , whose product is  $\text{'}c\text{'}$ ( the last term )  and whose sum is  $\text{'}b\text{'}$ ( coefficient of middle term).

 Rewrite the trinomial as  $x^2+ rx + sx + c$   and then use grouping and the distributive property to factor the polynomial. The resulting factors will be $\left(x + r\right)\text{ and }\left(x + s\right).$

Calculation:

In order to factor the given polynomial $2x^3+4x^2-8x$ , factor out the GCF from all terms and then simplify further by splitting the middle term in two such factors whose product is equal to the product of the first and third term. And then simplify further by grouping the terms and taking GCF of each group as shown below

  $\begin{array}{c}2x^3+4x^2-8x=2x\left(x^2+2x-4\right)\\ =2x\left(x^2+2x-4\right)\end{array}$

Thus, the given expression in factored form is $2x\left(x^2+2x-4\right)$ .