Algebra 2 help (Factoring)?

Question

Ok so I don't understand how to do these...I can do regular factors like x2+31x+30 but I need help with the following::
6x2+23x+6 10x2+23x+6 16x2-56x+49

Answer 1

x2 + 31x + 30, you find 2 numbers that multiply to 30 and add to 31..

10x2 + 23x + 6, you need to multiply the 1st and last #s (10 x 6 = 60). So you need to find 2 numbers that multiply to 60 and add to 23.

The #s are 20 and 3. Split up your middle term into these 2 #s.

10x2 + 20x + 3x + 6.

Look at the first 2 terms. What do they have in common? Take that out. Same with the last 2.

10x(x + 2) + 3 (x + 2).

If you do it right, the thing in the parentheses shoud be the same for both. Now use the distributive property backwards to factor:

(x + 2) (10x + 3)

If you can't factor, use the quadratic formula.

Answer 2

Don't know why the previous answerer thinks this is the wrong place... it is mathematics, after all.

6x^2 + 23x + 6: discriminant is 23^2 - 4(6)(6) = 529 - 144 = 385 which is not a square number, so there are no nice factors. Use the quadratic formula to get the solutions.

10x^2 + 23x + 6: We need to look for factors of 10 and factors of 6 that can combine to make 23. If we break 10 into 10 and 1 and 6 into 2 and 3 we can make 10*2 + 1*3 = 23. So the factorisation is
(10x + 3)(x + 2)

16x^2 - 56x + 49: As a shortcut, note that 16 = 4^2, 49 = 7^2 and 56 = 2*4*7. So this is just (4x - 7)^2.

Answer 3

FACTORISING!! NOT "Factoring". Now that's off my chest...

There are many ways to factorise. Mostly with hard ones like you had there I use the quadratic formula:
x= (-b ± √b^2 - 4ac) / 2a
where in the original form it is ax^2+bx+c

so for the first one:
6x^2+23x+6
a=6
b=23
c=6

Just plug those numbers into the quadratic formula to find out the two values for x. The first one is x= (-b+ √b^2 - 4ac) / 2a and the second, x= -b - √b^2 - 4ac) / 2a.

I hope that helped you.

Answer 4

These are the same idea, just a bit more complex
(6x^2+23x+6)
(6x+6)(x+1), is 6x^2+12x+6, so, nope
(6x+1)(x+6), is 6x^2+37x+6, so nope
(6x+3)(x+2), 6x^2+15x+6, nope
(6x+2)(x+3), 6x^2+20x+6, nope
(3x+1)(2x+6), 6x^2+18x+2x+6, nope
(3x+6)(2x+1), 6x^2+13x+6, nope
(3x+3)(2x+2), 6x^2+12x+6, nope
(3x+2)(2x+3), 6x^2+13x+6, nope

This is not factorable

10x^2+23x+6
(5x+1)(2x+6) = 10x^2+32x+6, nope
(5x+6)(2x+1)=10x^2+13x+6, nope
(5x+3)(2x+2)=10x^2+16x+6, nope
(5x+2)(2x+3)=10x^2+19x+6, nope
(10x+1)(1x+6)=10x^2+61x+6, nope
(10x+6)(x+1)=10x^2+16x+6, nope
(10x+2)(x+3)=10x^2+32x+6, nope
(10x+3)(x+2)=10x^2+23x+6, finally

Do the same for the last.

Answer 5

(n-8) ------- is the simplified type (n+8) n^2 -sixty 4 could be factored into (n+8)(n-8) (n+8)^2 is the comparable as (n+8)(n+8) so which you in simple terms divide with the aid of with the aid of the essential component to (n+8) and you get the asnwer i reported approximately.

Answer 6

Combine like terms and then use the Quadratic Formula

Answer 7

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut17_quad.htm

The above website has the basic formulae and some examples that you may want to read up on.