Find the real-number solutions of the equation

I"m confused as to how to begin to solve these (polynomials). Please help.

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!

Re: Find the real-number solutions of the equation

Quote:

Originally Posted by **dtbrunson**

I"m confused as to how to begin to solve these (polynomials). Please help.

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.

[tex]x^3+7x^2+4x+28=0[/tex]

Group the first two terms and factor out [tex]x^2}[/tex]

Then, group the last two terms and factor out a 4.

[tex]x^2(x+7)+4(x+7)=0[/tex]

Now you have: [tex](x^2+4)(x+7)=0[/tex]

Looks like 1 real and 2 imaginary solutions.

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)?.......

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)?.......

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:

[tex](x^2+4)(x+7)=0[/tex]

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

[tex]x^2+4=0[/tex]

[tex]x^2=-4[/tex]

[tex]x=\pm 2i[/tex]

These are your two imaginary zeros.

[tex]x+7=0[/tex]

[tex]x=-7[/tex]

This is your one real zero.

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...