Factoring polynomials?

Question

completely forgot how to factor polynomials. :(
can anyone help me?

b^3 + 6b^2

s^3 + 6s^2 + 11s

2x^3 + 4x^2 - 8x

Answer 1

The first step in any factoring situation is to look for some common term that can be factored out of all the terms. We sometimes call it the Greatest Common Monomial Factor ("monomial" is another word for a single term).

In the first case, each term has a factor of b², so we factor that out. Inside the parentheses, we write down the result of dividing each of the original terms by that factot. b³ / b² = b, so the first term will be b. 6b² / b² = 6, so our second term will be 6. So our answer is b²(b + 6).

You can test the answer by multiplying the b² back into the (b+6) using the distributive property. Remember, factoring is just a matter of breaking a product (a multiplication) down into its component pieces; when you're done, you can always multiply them back together to check.

I won't do the other two, but I'll get you started: the common factor in the 2nd one is s, and in the 3rd one, it's 2x.

Hope that helps!

Answer 2

b^3 + 6b^2
This one is pretty simple. Just factor out a 'b^2' from the equation and you've solved it
b^3 + 6b^2 = b^2 (b + 6)

s^3 + 6s^2 + 11s
First factor out an 's' from the equation. Upon further inspection, the equation cannot be factored further.
s^3 + 6s^2 + 11s = s (s^2 + 6s + 11)

2x^3 + 4x^2 - 8x
First factor out an '2x' from the equation. Upon further inspection, the equation cannot be factored further.
2x^3 + 4x^2 - 8x = 2x (x^2 + 2x - 4)

Answer 3

Step 1) pull out common factors. Each of your problems has a common factor. it is b^2 in the first one. s in the 2nd and 2x in the third.

Step 2) For trinomials, try guess and check or use the sum and product rule. In number 2 you would be looking for two numbers that multiply to give 11 and add to give 6. There aren't any.

Your answers are:

b^2(b+6)

s(s^2 + 6s + 11)

2x(x^2 + 2x -4)

These do not factor any further over the rationals. The second has complex roots and the 3rd has a root involving sqrt(20)

Answer 4

1. b^3 + 6b^2
= b.b^2 + 6.b^2
= b^2 (b+6)

2. s^3 +6s^2 + 11s
= s(s^2 + 6s + 11)

3. 2x^3 + 4x^2 -8x
= 2x(x^2 + 2x - 4)

Answer 5

(1)
b^3 + 6b^2 => (b^2)(b+6)

(2)
s^3 + 6s^2 + 11s => s(s^2 + 6s + 11)

(3)
2x^3 + 4x^2 - 8x => 2x(x^2 + 2x - 4)


Doug

Answer 6

b^2(b+6)

s(s^2+6s+11)

2x(x^2+2x-4)

http://www.quickmath.com

Answer 7

Factoring polynomials completely: x^3y + 2x^2y^2 + xy^3 Factor out the Greatest Common Factor (GCF) = xy xy(x^2 + 2xy +y^2) Factor a trinomial xy(y + x) (y +x) Final result: xy(y + x) (y + x)

Answer 8

1) all terms have b^2 common....
so

(b^2)(b+6)

2) as before...
s(s^2 + 6s + 11)
the quadratic given cannot be factorised because its roots are imaginary.

3) as before....
2x(x^2 + 2x - 4)
using quadratic formula.... (-b+_(b^2 - 4ac)^0.5)/(2a)

2x(x - (-2+(2^2 - 4*1*-4)^0.5)/(2*1))(x - (-2-(2^2 - 4*1*-4)^0.5)/(2*1))
calculating...

2x(x + 1 - 5^0.5)(x + 1 + 5^0.5)

Answer 9

b³ + 6b²

b² (b + 6)

Common factor is b²

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

s³ + 6s² + 11s

s (s² + 6s + 11)

Common factor is " s "

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

2x³ + 4x² + 8x

2x (x² + 2x + 4)

The common factor is 2x

Answer 10

Factoring is the "F word" of mathematics. Let me see if I can help you out.