Expert: Josh - 1/27/2004

Question
Solve by factoring.

1. 4x^2-15x=4

2. 12x^2-11x-15=0

3. 3x^2-48x=0

4. x^2=3x+10

Answer
Sandra,

Rather than showing you how to do a couple of questions, you really need to learn the principles.

Here is the general strategy.
(a) manipulate the equation into the standard form.
In general, let "a","b","c" be some numbers. We use these to represent the coefficients for the x^2,x and constant term respectively.
The standard form of the quadratic equation is,
a*x^2+b*x+c=0.  ...[#1]
(b) then, factorize the quadratic.
Consider the following observation.
Suppose that x=p, x=q are the two solutions.
then, we must have (x-p)(x-q)=0.
ie., x^2-(p+q)x+pq=0 ...[#2]
This is because, when x=p OR x=q, the equation yields zero, and we would have solved the equation, right?
So, if we divide the [#1] throughtout by "a"
x^2+(b/a)*x+(c/a) must be identical to x^2-(p+q)x+pq.
Since both are equal to zero.

Okay, how do we reconcile all these information.

Q4.,
[Step a] Manipulate x^2=3x+10 into the standard form of the quadratic equation, a*x^2+b*x+c=0 as in [#1]
[Working a]
x^2-3x-10=0 Okay?
ie., for this question, a=1,b=-3,c=-10.

[step b] using the argument discussed above, if we let x=p,x=q be the solutions, then, x^2+(b/a)*x+(c/a) must be identical to x^2-(p+q)x+pq. Since both are equal to zero.

So, x^2-3x-10=x^2-(p+q)x+pq.

[step c] comparing the coefficients,
For the x term: p+q=3, ...[#a]
For the constant term, pq=-10. ...[#b]

-10 is the product of two numbers, in particular, -10=5*(-2)
As, you can see, 5+(-2)=3 satisfies [#a]
so we have worked out the factors, here, p=5,q=-2.

That is, x^2-3x-10=(x-5)(x+2)=0 and x=5,-2.

-----
More generally, let m,n,r,s be any number.
we want the standard equation to be factorized as (mx-n)(rx-s)=0.
That is, we want the standard equation to appear as
mr*(x^2)-(ms+nr)*x+ns=0

For Q1, we have 4x^2-15x=4 or 4x^2-15x-4=0

by inspection, observe that 4*(-4)=-16 and the middle coefficient -15 may result from adding 1 to -16.
Very quickly, we speculate that (4x-n)(x-4)=0 ...[#c]
By letting n=-1, it works out.

Formally, comparing (mx-n)(rx-s)=0. with 4x^2-15x-4=0
we have mr*(x^2)-(ms+nr)*x+ns=4x^2-15x-4
ie.,
For x^2: mr=4
For x^1: ms+nr=15
For x^0: ns=-4

It is easiest to solve this by inspection.
If we let m=4, r=1, s=4 as we speculated in [#c]
Then for x^1, ms+nr=16+n=15 and n=-1.

To summarize,4x^2-15x-4=(4x+1)(x-4).
You work out the solutions by setting each linear factor to zero. That is, (4x+1)=0 when x=-1/4, (x-4)=0 when x=4.
So, the solutions are x=-1/4, x=4.
I think you would agree that solving this question by inspection is the easiest way to go.

Q3, 3x^2-48x=0
pull out common factor,
3x(x-16)=0 when 3x=0, i.e., x=0 OR
when x-16=0, i.e., x=16.

Q2., Setting 12x^2-11x-15=0
Consider 12 as the product of two numbers.
The factors of 12 are {1,12,2,6,3,4}
Consider 15 as the product of two numbers.
The factors of 12 are {1,15,3,5}
After trial and error,
we find that by setting mr*(x^2)-(ms+nr)*x+ns=12x^2-11x-15
ie.,
For x^2: mr=12
For x^1: ms+nr=11
For x^0: ns=15

Letting m=4,r=3,
n=-3,s=5
ms=-20,nr=9 and ms+nr=11 and it works out.

So, the factorized form is (mx-n)(rx-s)=(4x+3)(3x-5)=0

solutions are x=-3/4,5/3.

Write back if anything is unclear.