How do you factor the quadratic equation #12x²+13x-4#?

In general, polynomials of any power
#x^n+a_1x^(n-1)+...+a_(n-1)x+a_n#
can be factored into a product of linear polynomials
#(x-b_1)(x-b_2)...(x-b_n)#
using their roots #b_1,b_2,...,b_n#, that is values of an unknown #x# that make the polynomial to evaluate to zero:
#b_i^n+a_1b_i^(n-1)+...+a_(n-1)b_i+a_n=0# for any #i=1,2,...,n#

In particular, a quadratic polynomial
#x^2+a_1x+a_2#
can be factored into a product of two linear polynomials
#(x-b_1)(x-b_2)#,
where #b_1# and #b_2# are solutions of an equation
#x^2+a_1x+a_2=0#

If there is a coefficient not equal to #1# at the highest power of an unknown #x#, we have to factor it out prior to the above transformations.

So, the straight forward method to factor this quadratic polynomial is to factor out #12# and to find two roots of the remaining polynomial, that is two solutions to an equation
#x^2+(13/12)x-4/12=0#

We can use a formula for solutions of this equation:
#x_(1,2)=(-13/12+-sqrt(13^2/12^2+4*4/12))/2=(-13+-19)/24#
From this we have two solutions:
#x_1=1/4# and #x_2=-4/3#

That brings the following factorization:
#12x^2+13x-4=12(x-1/4)(x+4/3)#

A quadratic expression is completely factorizable if and only if its discriminant is positive. Given a quadratic expression of the form #ax^2+bx+c#, the discriminant #\Delta# is defined as #b^2-4ac#. In your case, we have #a=12#, #b=13# and #c=-4#. For this values, we have #\Delta=361#, which means that we can factor the expression finding two solutions #x_1# and #x_2#, and thus writing #12x^2+13x-4=(x-x_1)(x-x_2)#.

To find the solutions, we have the formula
#x_{1,2}=\frac{-b\pm \sqrt(\Delta)}{2a}#.
Since #\Delta=361#, its square root equals 19. Plugging the values, we have the two solutions
#x_1={-13+19}/{24}=1/4# and
#x_2={-13-19}/{24}=-4/3#.

The factorization is thus #(x-1/4)(x+4/3)#

#12x^2 +13x-4 = (4x-1)(3x+4)#

Read though the method carefully.
It's worth the time spent to learn it.

Explanation:

#12x^2 +13x-4# is an example of a quadratic (#x^2#) trinomial.(3 terms)

Some trinomials are the product of two binomials of the form:

#(x+-a)(x+-b)#

When we try to find these factors from the quadratic, we are factoring or factorising. Not all quadratic trinomials can be factored.

In #color(magenta)(12)x^2 +color(lime)(13)x-color(magenta)(4)#

We are trying to find factors of 12 and 4 which subtract to make 13.
Look at the clues.

CLUE 1 : The sign with the 4 is negative

#rarr# The signs in the the brackets are different
#color(white)(xxxxxxx)(+) xx( -) rarr ( -)#
#rarr# The difference between the factors must be 13.

CLUE 2: 13 is an odd number, It can only be obtained from subtracting with an odd and an even number.

The factors of 4 are # 1xx4 or cancel(2xx2)#
In this case it will not be #2xx2# because that will make both pairs even.

#rarr#The required factors of 4 are #1 and 4#

The factors of 12 are: #1xx12 and cancel(2xx6) and 3xx4#
#2 and 6# are both even - reject them as explained above.

Thinking through the possible combinations might go like this:

12x4 = 48 ....too big
12x1 = 12........too small - we still need to subtract.

Let's try 4 and 3 with 1 and 4. Cross-multiply and subtract

#color(white)(x.) (12)" "(4)#
#color(white)(x.x) darr" "darr#
#color(white)(xxx) 4" "1 rarr 3xx1 = 3#
#color(white)(xxx) 3" "4 rarr 4xx4 = 16 " subtract: " 16-3 = 13#

#rarr#We have the correct factors!! Now for the signs.

#+13 rarr# more positives. #color(red)(-3)color(blue)( +16) = +13#
Make them #color(red)(-3) and color(blue)( +16)#

#color(white)(xxxxxx)darr# Insert the signs here between the factors.

#color(white)(xxx) 4" "color(red)(-1)rarr 3xxcolor(red)(-1) = color(red)(-3)#
#color(white)(xxx) 3" "color(blue)(+4) rarr 4xxcolor(blue)(+4) = color(blue)(+16)#

The top row has the factors for the first bracket.
The bottom row has the factors for the second bracket.

#12x^2 +13x-4 = (4x-1)(3x+4)#

You can check by multiplying out again.

This requires a good deal of practice and a solid knowledge of the multiplication tables. However, once mastered it is a very quick method which works most of the time.

It does avoid having to find all the possible factors of numbers which are sometimes quite big.