Help please? I need to solve by factoring.

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Answer 1 · 2017-11-30T06:18:37

Sure I'll help.

In order to factor the equation, we have to arrange the equation so it is all multiplication (essentially).

$2x^4 + 12x^3 + 16x^2 = 0$

First let's factor out the $2x^2$ .

$2x^2(x^2+6x+8)=0$

Now, we are left with a simple trinomial in the brackets. Factor it normally.

$2x^2(x^2 + 4x+2x+8)=0$

$2x^2(x[x+4]+2[x+4])=0$

$2x^2(x+2)(x+4)=0$

Now that we have fully factored the equation, we can use this to find the solutions.

The solutions are points on the graph when $y=0$ . To get this, $x$ needs to equal to something that makes the entire equation $0$ .

Let's start with $2x$ . To get $0$ , anything multiplied by $0$ equals $0$ . Therefore, $x=0$ . That's one solution.

Next is $x+2$ . Well, to get $0$ here, $x$ has to be $-2$ , becasue $-2+2=0$ . That's the second solution.

Finally, $x+4$ . The same thing as before also applies here, except $x=-4$ .

Therefore, the solutions to the function are $x=0, -2, -4$ .

We can double check this by graphing the the function.

graph{2x^4 + 12x^3 + 16x^2 [-22.8, 22.82, -11.4, 11.4]}

Hope this helps :)