## Question

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

## Answers

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

subtractto make 13.

Look at the clues.

CLUE 1: The sign with the 4 isnegative

#rarr# The signs in the the brackets aredifferent

#color(white)(xxxxxxx)(+) xx( -) rarr ( -)#

#rarr# Thedifferencebetween 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 willnotbe#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.