New problem same topic

[PuppyGirl]

Directions: Find all real-number awnsers.

x^3+3x^2-2x-6=0
(x^3+3x^2)(-2x-6)=0
x^2(x+3)-2(x-6)=0
Hint: a couple of awnsers are \/¯2, -3. This is where I'm stuck x^2(x+3)-2(x-6)=0

How do I find the awnsers?

 

[stapel]

x^3+3x^2-2x-6=0
(x^3+3x^2)(-2x-6)=0

How did the subtraction, "x³ + 3x² - 2x - 6", become a multiplication, namely "(x³ + 3x²)(-2x - 6)"?

(x^3+3x^2)(-2x-6)=0
x^2(x+3)-2(x-6)=0

How did the multiplication turn back into subtraction? How did you get that (-2)(-6) = -6?

How do I find the awnsers?

If "answers" is your book's term for "solutions", "roots", or "zeroes", then use the techniques from class: the Rational Roots Test, synthetic division, etc. For instance, the Rational Roots Test would give you the list "±1, ±2, ±3, ±6" as possible roots. Then you'd plug these, one by one, into the synthetic division algorithm to find one or more that worked.

Or are you supposed to be using some other method for solving the polynomial? Or does "answers" means something different?

Please reply with clarification. Thank you.

Eliz.

 

[PuppyGirl]

I don't know the Rational Roots Test, or Syntheic division etc. I only know how to factor a polynomial. If you could help me out with these problems by factoring polynomials that would be great.

Thank You,

Heidi T.

 

[Mrspi]

Directions: Find all real-number awnsers.

x^3+3x^2-2x-6=0
(x^3+3x^2)(-2x-6)=0
x^2(x+3)-2(x-6)=0
Hint: a couple of awnsers are \/¯2, -3. This is where I'm stuck x^2(x+3)-2(x-6)=0

How do I find the awnsers?

Since you said you have not studied the Rational Roots Theorem, or synthetic division, I wonder if you have tried factoring the left side.

If a polynomial expression has more than 3 terms, you may find it helpful to group the terms. Let's group the first two terms together, and the last two terms together:

x<SUP>3</SUP> + 3x<SUP>2</SUP> - 2x - 6

Remove a common factor of x<SUP>2</SUP> from the first two terms, and remove a common factor of -2 from the last two terms:
x<SUP>2</SUP>(x + 3)[/color] - 2(x + 3)

Ok.....now, do you see that (x + 3) is a common factor? Remove it, and you have
(x + 3)(x<SUP>2</SUP> - 2)

And the equation is now
(x + 3)(x<SUP>2</SUP> - 2) = 0

Set each factor equal to 0, and solve.