Factor. Look for a GCF first. X^3+2x^2-48x

Mathematics

1 answer:

svp [43]1 year ago

4 0

Answer:

x(x - 6)(x + 8)

Step-by-step explanation:

First, the GCF of these 3 terms is x.

So to separate x from this it now becomes x(x^2 + 2x - 48).

Now to factor the expression inside the parentheses easily I use "a" and "b" to make it easier.

a + b = 2 (from 2x)

a * b = -48 (from third term of expression)

a = -6, b = 8

Substitute "a" and "b" and I get: x(x - 6)(x + 8)

You might be interested in

Express the quotient of z1 and z2 in standard form given that <img src="https://tex.z-dn.net/?f=z_%7B1%7D%20%3D%20-3%5Bcos%28%5C

Lesechka [4]

Answer:

Solution : $-\frac{3}{4}-\frac{3}{4}i$

Step-by-step explanation:

$-3\left[\cos \left(\frac{-\pi }{4}\right)+i\sin \left(\frac{-\pi \:}{4}\right)\right]\:\div \:2\sqrt{2}\left[\cos \left(\frac{-\pi \:\:}{2}\right)+i\sin \left(\frac{-\pi \:\:\:}{2}\right)\right]$

Let's apply trivial identities here. We know that cos(-π / 4) = √2 / 2, sin(-π / 4) = - √2 / 2, cos(-π / 2) = 0, sin(-π / 2) = - 1. Let's substitute those values,

$\frac{-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)}{2\sqrt{2}\left(0-1\right)i}$