Factor by splitting the middle term 12x^2+17x+6?

12x^2+17x+6 = (12x^2+9x)+(8x+6)


=3x(4x+3)+2(4x+3)


=(3x+2)(4x+3) - PKFactor by splitting the middle term 12x^2+17x+6?
12x^2+17x+6=12x^2+8x+9x+6


=4x(3x+2)+3(3x+2)


=(3x+2)(4x+3) AnsFactor by splitting the middle term 12x^2+17x+6?
To factor this you need factors of 12 and factors of six that when multiplied and added = 17. The factors of 12 are 12, 6, 4, 3, 2, 1 and the factors of 6 are 6, 3, 2, 1. By trial and error you can get 12x^2 + 17x + 6 = (4x + 3)(3x + 2).





Proof: (4x + 3)(3x + 2) ==%26gt; 4x(3x + 2) + 3(3x + 2)


==%26gt; 4x(3x) + 4x(2) + 3(3x) + 3(2) ==%26gt; 12x^2 + 8x + 9x + 6


==%26gt; 12x^2 + 17x + 6
12x^2+9x+8x+6


3x(4x+3)+2(4x+3)


=(4x+3)(3x+2)
The trick here is to try to find a factor which is common to more than one grouping of terms.





If we split up the above expression as follows, and do a little thinking and paper scratch work, we see that there is a common factor:





12x^2+8x+9x+6 = (12x^2+8x)+(9x+6).





Now factor each grouping of terms separately.





(12x^2+8x) = 4x(3x+2)


(9x+6) = 3(3x+2)





Now all we have to do is put the parts back together again and extract the appropriate common factor(s).





12x^2+8x+9x+6 = 4x(3x+2)+ 3(3x+2) = (4x+3)(3x+2).





Notice that if we actually multiplied the last two factors out longhand, we will get the initial expression.
- short answer :


12x^2+17x+6=12x^2+(8x+9x)+6=(12x^2+8x)鈥?br>

- long answer if you wanna know more :


The example was built to be an easy one to be solved by splitting the middle term.


This is a degree 2 equation, which generally looks like this : f(x)=ax^2+bx+c. in your case a=12,





b=17 and c=6. However, the most common thing to do with an equation is to solve it, so we may say





: for what ';x'; f(x)=0 ? If f(x)=0, also g(x)f(x)=0, whoever g(x) is....


Solving the degree 2 equation by splitting the middle term is the easiest way to do it, but it is





not always possible to split.... here, split this : x^2+1.416666x+0.5


Believe me, this is ';THE SAME EXAMPLE';, so what is beyond splitting ??


First you must know how to solve a degree 2 equation that generally is f(x)=ax^2+bx+c; you follow





this steps:


1. calculate delta=b^2-4ac; for you : delta=17^2-4x12x6=289-288=1


2. calculate sqrt(delta) if delta%26gt;=0. If delta is negative you are dealing with complex numbers,





but that's another story......so sqrt(delta)=sqrt(1)=1, and -1. This is something that few people





know: there are 2 square roots for one given number, and this roots are equal in absolute value,





but they differ by sign. Also sqrt(9)=3 and -3, because (-3)^2=9 get it?.......


3. if delta%26gt;=0, than you calculate the values ';x'; for which f(x)=ax^2+bx+c=0, just like this :





x=(-b+sqrt(delta))/(2a), which will generate 2 values(REMEMBER!! because sqrt(delta) generates 2





values), so x=(-17+sqrt(1))/(2x12); first value is (-17-1)/24=-18/24=-3/4; second value is





(-17+1)/24=-16/24=-2/3. This 2 values are x1 and x2, also known as the roots of f(x).


4. if you have the roots, you may wright f(x) like this : f(x)=a(x-x1)(x-x2). so





12(x-(-2/3))(x-(-3/4))=12(x+2/3)(x+3/4鈥?br>

Now remember when i sad if f(x)=0, also g(x)f(x)=0 ?? please observe that if x=x1 or x=x2, your





equation is 0(zero), because one or another of those parenthesis (x+2/3)or(x+3/4) goes to 0. So





that ';12'; doesn't matter. So you may put whtvr number you wanna in the place of 12. The example i





asked you to ';split'; : x^2+1.416666x+0.5 is actually your example without the ';12';.


Now if you remember, i sad ';The example was built to be an easy one to be solved by splitting the





middle term'; : if you sum your roots by fractional laws: x1+x2=2/3+3/4 you must multiply the 1st





one with 4, and the 2nd with 3, and you will have : 8/12+9/12=17/12. Do you see now: 17 is the





term to split....and the entire equation was multiply by ';12'; so you don't have to split a





fraction.So this is how you built ';easy to split'; examples.


Going even Dipper, if x1 and x2 are the roots of ax^2+bx+c, x1+x2=-b/a and x1*x2=c/a, so ';b'; is





kind of their sum, and ';c'; is kind of their prod....These are the ';Viette'; relations....
(3x + 2)(4x + 3)