factoring

[mafrediani]

Ok this may be very basic but can anyone tell me how to factor 2n + 9n -56 The 2n should be squared (i didnt know how to type it)

 

[Yogi]

Factor 2n^2 + 9n -56:

Often, basic trinomials can be factored into the product of 2 binomials.

(___ ___)(___ ___)
You just need to fill in the blanks and the signs to make it work with FOIL.

 

[Lost souls]

Ok this may be very basic but can anyone tell me how to factor 2n + 9n -56 The 2n should be squared (i didnt know how to type it)

In case of quadratic equations of the form Ax^2+Bx+C=0,
a+b = B
a*b= A*C
u may use trial and error method to find a and b.
the equation would be
Ax^2+ax+bx+C=0


if coefficient of x² is 1,
eqn can be written as (x+a)(x+b)=0


2n²+9n-56=0
2n²+16n-7n-56=0
2n(n+8)-7(n+8)=0
(2n-7)(n+8)=0

 

[Willie]

Here's another approach:

Original expression: 2n^2 + 9n -56

Multiply the leading coefficient by the last term: 2*(-56)=-112

Find two factors of this result that will add to the coefficient of the middle term. In this case, -7 and 16 will work. (-7*16=-112 and -7+16=9)

Now use these two factors to create two new middle terms to replace the original one: 2n^2 + 16n -7n -56 Notice that the two middle terms (16n and -7n) can be combined to 9n, so this expression is equivalent to the original one.

Now factor this expression by grouping:

(2n^2 +16) + (-7n - 56) <-- group into two binomials
2n(n + 8) -7(n+8) <-- extract common monomial factor from each of the binomials
(2n-7)(n+8) <-- apply distributive law in reverse

All done.

Edit:

If you FOIL the result when you use this method, that will serve not only to check your answer, but will also give you some understanding of why this method works.

 

[lookagain]

In case of quadratic equations of the form Ax^2+Bx+C=0,
a+b = B
a*b= A*C
u may use trial and error method to find a and b.


the eqn is factored down to (x+a)(x+b)=0. **

This isn't true. Here, a + b = 9, ab = 2(-56) = -112.

So, either a = -7 and b = 16, or a = 16 and b = -7.

Then, ** would be (x - 7)(x + 16) = 0, or (x + 16)(x - 7) = 0.

The solutions of either of those equations ** are 7, -16.

The solutions of the OP's trinomial set equal to 0 are -8, 7/2.





hence , 2n²+9n-56=0

2n²+16n-7n-56=0
2n(n+8)-7(n+8)=0
(2n-7)(n+8)=0

...

 

[HallsofIvy]

In case of quadratic equations of the form Ax^2+Bx+C=0,
a+b = B
a*b= A*C
u may use trial and error method to find a and b.
the eqn is factored down to (x+a)(x+b)=0.

hence , 2n²+9n-56=0
2n²+16n-7n-56=0
2n(n+8)-7(n+8)=0
(2n-7)(n+8)=0

Interesting! You got the correct answer but did NOT use the method you give. In particular, (2n- 7)(n+ 8) is NOT of the form "(x+a)(x+ b)".

mafediani, While any quadratic can be (theorectically) factored, there are simply too many possibilities unless the coefficients are integers. So, normally, when we talk of "factoring" a polynomial we mean finding integers a, b, c, and d so that

(ax+ b)(cx+ d)= acx^2+ (bc+ ad)x+ bd= Ax^2+ Bx+ C so we must have ac= A, bc+ad= B, and bd= C. That is three equations to find 4 numbers so unless we put restrictions on the numbers, there are just too many things to try. (And, unfortunately, almost all polynomials can't be factored with integer coefficients.)

Here, we are looking for two integers, a and c, such that ac= 2. Fortunately, there are only two ways to do that: a= 1, c= 2 or a= 2, c= 1 (and which we use doesn't really matter). There are many numbers, b and d, such that bd= -56= 2(-28)= (4)(-14)= 8(-7). (or the same numbers with the signs reversed) But we also want bc+ ad= 2(b)+ 1(c)= 9. And the quick way to find the right pair (if it exists) is to try each: with b= 2, d= -28, 2(2)+ 1(-28)= -24; with b= 4, d= -14, 2(4)+ 1(-14)= -8; with b= 8, d= -7, 2(8)+ 1(-7)= 9!! So we have found that a= 1, b=8, c= 2, and d= -7 which gives exactly the answer Lost Souls did- (x+ 8)(2x- 7).

That's a bit more complicated that Lost Souls indicated because the coefficient of $\displaystyle x^2$ is NOT 0. And notice that, as is often the case in mathematics, a lot of it is trial and error!

 

[HallsofIvy]

When "difficult", call on the quadratic equation.
Google it. Much better if you learn by yourself.

I get terribly embarassed when I use the quadratic formula and then find out that the roots are simple integers!

 

[mmm4444bot]

I get terribly embarassed when I use the quadratic formula and then find out that the roots are simple integers!

Check the Discriminant, first. If this does not provide relief, I can recommend some drugs.

 

[JeffM]

I get terribly embarassed when I use the quadratic formula and then find out that the roots are simple integers!

I am with denis on this one. If I do not "see" a factorization quickly, I stop messing around and use the quadratic formula. I may be using a sledge hammer to crack a peanut, but I am not wasting time. I do not find efficiency embarrassing. Of course, having been trained academically as an historian, my lack of mathematical "vision" is natural.