Find the real-number solutions of the equation

[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!

 

[masters]

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.

\(\displaystyle x^3+7x^2+4x+28=0\)

Group the first two terms and factor out \(\displaystyle x^2}\)
Then, group the last two terms and factor out a 4.

\(\displaystyle x^2(x+7)+4(x+7)=0\)

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

Looks like 1 real and 2 imaginary solutions.

 

[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)?.......

 

[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)?.......

 

[masters]

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:

\(\displaystyle (x^2+4)(x+7)=0\)

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

\(\displaystyle x^2+4=0\)

\(\displaystyle x^2=-4\)

\(\displaystyle x=\pm 2i\)

These are your two imaginary zeros.

\(\displaystyle x+7=0\)

\(\displaystyle x=-7\)

This is your one real zero.

 

[Denis]

(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...