# Find the real-number solutions of the equation

• 05-14-2010, 03:32 PM
dtbrunson
Find the real-number solutions of the equation

x^3 +7x^2 +4x +28 = 0

I don't know how to type these problems out on the computer, so I hope you understand. Here's where I started and then got lost.

x(x^2 + 7x + 4....... that s as far as I got. Please help it makes no sense to me!
• 05-14-2010, 03:56 PM
masters
Re: Find the real-number solutions of the equation
Quote:

Originally Posted by dtbrunson

x^3 +7x^2 +4x +28 = 0

I don't know how to type these problems out on the computer, so I hope you understand. Here's where I started and then got lost.

x(x^2 + 7x + 4....... that s as far as I got. Please help it makes no sense to me!

Hi dtbrunson,

You didn't notice you could factor this expression by grouping, did you?
Well, you can.

$$x^3+7x^2+4x+28=0$$

Group the first two terms and factor out $$x^2}$$
Then, group the last two terms and factor out a 4.

$$x^2(x+7)+4(x+7)=0$$

Now you have: $$(x^2+4)(x+7)=0$$

Looks like 1 real and 2 imaginary solutions.
• 05-14-2010, 04:09 PM
dtbrunson
Re: Find the real-number solutions of the equation
Thanks but what do you mean when you say 1 real and 2 imaginary? Does it go further than that?
(x^2 + 4) (x + 7)
x(x+4) (x+7)?.......
• 05-14-2010, 08:56 PM
dtbrunson
Re: Find the real-number solutions of the equation
Thanks but what do you mean when you say 1 real and 2 imaginary? Does it go further than that?
(x^2 + 4) (x + 7)
x(x+4) (x+7)?.......
• 05-17-2010, 12:44 PM
masters
Re: Find the real-number solutions of the equation
Quote:

Originally Posted by dtbrunson
Thanks but what do you mean when you say 1 real and 2 imaginary? Does it go further than that?
(x^2 + 4) (x + 7)
x(x+4) (x+7)?.......

That's factored incorrectly. Start here:

$$(x^2+4)(x+7)=0$$

Use the 'zero product property' to set each factor to zero and solve.

$$x^2+4=0$$

$$x^2=-4$$

$$x=\pm 2i$$

These are your two imaginary zeros.

$$x+7=0$$

$$x=-7$$

This is your one real zero.
• 05-17-2010, 12:55 PM
Denis
Re: Find the real-number solutions of the equation
Quote:

Originally Posted by dtbrunson
(x^2 + 4)
x(x+4)

NO. x(x + 4) = x^2 + 4x, NOT x^2 + 4.

And if you don't understand what masters just told you, then you need help ftom your teacher...