Question

How do you factor `3c ^ { 2} + 24c + 45`?

Answer 1

This should factor down to `3(c+3)(c+5)`

Explanation:

First, let's look for general multiples:
`3c^2+24c+45`

all 3 coefficients (3, 24, 45) have a common divisor of 3. Let's pull that out:

`3(c^2+8c+15)` (eqn. 1)

Now, we can work inside the parentheses and break down the equation:

`3(c+x)(c+y)`, where:

`x+y=8` (eqn. 2)
and
`x*y=15` (eqn. 3)

we know this because expansion shows:
`(c+x)(c+y)=c^2+(x+y)c+xy`

Now we can solve for x & y two ways: guessing and direct solution. The direct solution will involve using the quadratic formula.

if you back up to equation 1, you can use the quadratic formula:
`(-b+-sqrt(b^2-4ax))/(2a)`
with:
a=1
b=8
x=15

Solving for this gives you the 'zeros' for the factored equation, which are c=-3 and c=-5. That means that if either value of c is plugged into eqn. 1, the answer is 0. You can then figure out that:

`c+x=0` and `c+y=0`, meaning x and y must be +3 and +5.

This gives you the final factored formula:

`color(red)(3(c+3)(c+5))`

The guessing method involves the use of eqn. 2 and eqn. 3. we need to figure out combinations of integer values that satisfy both equations.

We know that both the sum and the product of the two numbers are positive, therefore making x and y positive. This leaves us with a number of solutions for eqn. 2, but only 2 solutions for eqn. 3: [1, 15] and [3,5]. Since 1+15 is twice as large as 8, it must be
`color(red)("3 and 5")`