6n^2-11n-2 Factoring Polynomials.?

Question

I don't really care the answer to the question, I more care about how to solve it. Please explain fully and explain this like you would to someone who has no idea how to do this. My teacher sucks, so idk how to do these.
6n^2-11n-2

Answer 1

Well, you can always use the quadratic formula but that is kind of a last resort.

Most people would solve it by guess and by gosh.

The factors would have to be

(6n + something) x (n + something)
or
3n + something) x (2n + something)
At this point it helps to know how to multiply two bi-nomials together because what you do next is try putting numbers in for the somethings.
Look at
(3n + something) x (2n + something)
The product of the somethings has to be -2. so the possibilities are
(3n + 2) (2n - 1) the cross product term is 4n - 3n = n which isn't equal to -11n.

(3n - 2) (2n + 1) cross product term = -n
(3n + 1) (2n - 2) cross product term = -4n
(3n - 1) (2n + 2) cross product term = 4n.

so let's try

(6n + something) x (n + something)

Again something x something = -2
and the possibilities are
(6n + 2) (n - 1) cross product = -4n
(6n - 2) (n + 1) cross product = 4n
(6n - 1) (n + 2) cross product = 11n
heart goes pitter pat
(6n + 1) ( n - 2) cross product = -11n
Ureka

The way that you develope this technique is practice, practice, practice.
And there is no law of God or man that says the factors have to involve easy integers. Sometimes the quadratic formula is the only way.
It helps if you can multiply bi-nomials easily.

Answer 2

This expression is a Quadratic, so it should factor into two linear factors in the form an+b and cn+d such that (an+b)(cn+d)=6n^2-11n-2.

By expanding, we see that ac = 6, bd = -2 and ad+bc=-11.

Experimenting with integer values shows that a=6, d=2, b=-1, and c=-1 works as an answer, ergo (6n+1)(n-2)=6n^2-11n-2... which checks correctly.

Alternately, you could set 6n^2-11n-2 equal to 0 and use the quadratic formula to find that the roots are 1/6 and -2. Setting n=1/6 or n=-2, we can algebreically show that 6n-1=0 or that n+2=0.

Multiplying these two equalities together shows that (6n+1)(n-2)=0 and we already assumed that 6n^2-11n-2=0 so (6n+1)(n-2)=0=6n^2-11n-2.

Either way, you show that 6n^2-11n-2=(6n+1)(n-2)

Answer 3

Remember how to multiply binomials?
multiply the like terms. Then the inside cross products and
outside cross products. add up to the middle term of the
polynomial. You can only get -2 by +1*-2 or -1*+2
Plus 6n^2 you can get a number of ways. So lets draw it out
( +1 )( -2) Since the factor of n is -11 lets just on a
hunch make the 2 negative Now what placement of the n's will
give us the inside and outside products we need to make
-11n and give us 6n^2 for the like product.
put 6x in the bracket with the +1. The outside product
6n*-2=-12n. Then put n in the bracket with the -2. The
inside product is n. Combine the like terms and you get
the desired -11n
answer. (6n+1)(n-2)

Answer 4

Brian, resist the temptation to cut yourself off from your teacher. Sometimes the crush of 25 or 30 students crimps his or her style, and the teacher might be more easy to work with at a different time. Now back to your question..........

This type of quadratic, where the squared term has a coefficient other than 1, is tricky because there is no closed-formed way to get an answer.
What ever method you use is basically trial and error with some "bells and whistles" on it. I think the method that is taught is to break up the polynomial into a form of
Ax *(Bx+C) + D*(Bx+C) or (Bx+C)(Ax+D)
From this, A*B=6
A*C+B*D= -11
and C*D=-2
A "reasonable" guess is (looking at the AC+BD=11 equation) A=6, C=-2, B=1 and D=1 or the breakdown is (x-2)(6x+1)

Answer 5

6n^2 - 11n - 2

this equation will help you graph pretty much any quadratic equation.

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-11) ± sqrt((-11)^2 - 4(6)(-2)))/(2(6))
x = (11 ± sqrt(121 + 48))/12
x = (11 ± sqrt(169))/12
x = (11 ± 13)/12
x = (24/12) or (-2/12)
x = 2 or (-1/6)

the problem will factor to (6x + 1)(x - 2)

if your wondering, its called a quadratic formula, its basically the conclusion to completeing the square.

Answer 6

ingredient out the easy 5x in each term first: 10x^3 - 5x^2 - 50x = 5x(2x^2 - x - 10) because of fact the 2x^2 is fundamental, a 2x is going in one ingredient and an x in the different: 5x(2x +- ?)(x +- ?) because of fact the ten is unfavorable, the signs and warning signs interior the climate are opposite: 5x(2x + ?)(x - ?) or 5x(2x - ?)(x + ?) Now you want 2 aspects of 10 that variety by ability of a million whilst certainly one of them is elevated by ability of two and the different by ability of a million. for the reason that 2 situations 2 is 4 and 5 situations a million is 5, the climate 2 and 5 are those to apply. set up them so as that the 5 multiplies by ability of x utilising the minus sign (to get the x term unfavorable). 5x(2x - 5)(x + 2)