Question

How do you factor the expression `5x^2 + 14x - 3`?

Answer 1

`y=(5x-1)(x +3 )`

Explanation:

We have

`y=5x^2 +14x -3`

We begin by multiplying `5` and `-3` to get `-15`. To continue we need two numbers that multiply to `15` and add to `+14`.

So we consider all whole number factors of `-15`
`-3,5` -> `-3+5=2`, no.
`3,-5` -> `3-5=-2`, no.
`1,-15` -> `1-15=-14`, no.
`-1,15` -> `-1+15=14`, yes!

We break up the `14x` term into `15x-1x` and put this in our function.
`y=5x^2 +14x -3` becomes

`y=5x^2 +15x -x -3`

Now we combine common factors from these four terms. We have one squared term, two linear terms and one constant term. Note we ALWAYS combine the square term with one of the linear terms and the other with the constant term, So this means there are only two ways to pair them up,

`5x^2` & `-x` and `15x` & `-3` or the other way
`5x^2` & `15x` and `-x` & `-3`.
I'm going to do it both ways so you're see how it works.

PAIRING ONE

Group our pairs.
`y=(5x^2 -x) + (15x -3)`

Now remove ANY common factors in the pairs, here `5x^2` and `-x` both have `x`'s we we factor out that and `15x` and `-3` both have a `3` in them so we factor out that. We are left with
`y=x(5x - 1) + 3(5x -1)`

Notice that the same term is left over from the factoring, `(5x-1)` and it is multiplying BOTH the `x` and the `+3`. So we factor this whole term out of both `x(5x - 1)` and `3(5x -1)`, leaving us with

`y=(x+3)(5x - 1)`

If you pair the other way it goes like this

`y=(5x^2 + 15x) + (-x -3)`
Removing common factors we have.
`y=5x(x +3 ) - 1(x + 3)`
Factoring out the `x+3` from the two remaining terms we have.
`y=(5x-1)(x +3 )` .

So the only difference is the order the terms fall out in, not what terms you get..

Answer 2

(5x - 1)(x + 3)

Explanation:

Use the new AC Method (Socratic Search)
`y = 5x^2 + 14x -3 =` 5(x + p)(x + q).
Converted trinomial: `y' = x^2 + 14x - 15 =` (x + p')(x + q').
p' and q' have opposite signs because ac < 0.
Factor pairs of (ac = - 15) --> (-1, 15). This sum is 14 = b. Then, p' = 1 and q' = 15.
Back to original y, `p = (p')/a = -1/5` and `q = (q')/a = 15/5 = 3`.
Factored form of y:
y = 5(x - 1/5)(x + 3) = (5x - 1)(x + 3).

Note: This new AC Method avoids the lengthy factoring by grouping and the solving of the 2 binomials.

Answer 3

`(5x-1)(x+3)`

Explanation:

All the information we need is in the trinomial itself.

`ax^2 +bx +c`

`color(lime)(5)x^2 +14x -color(lime)(3)`

We need to use factors of `color(lime)(5 and 3)larr` factors of a and c

`5x^2 +14x color(red)(-)3" "larr` look at the sign of c
`color(white)(xxxxxxx)uarr`

The `color(red)("MINUS")` sign of the third term tells us TWO things:

`rarr color(red)("SUBTRACT")` the factors of `5 and 3`

`rarr` the signs in the brackets will be `color(red)("DIFFERENT")`

`5x^2 +color(blue)(14)x -3`
`color(white)(xXxx)uarr`
`color(blue)(14)` is the answer when we subtract the factors of `5 and 3`

`5x^2 color(magenta)(+)14x -3" "larr` look at the sign of b
`color(white)(x.x)uarr`

The `color(magenta)("PLUS")` sign tells us there must be more positives.

In this case 5 and 3 are prime numbers, so there are only two combinations to try.

However we note that the biggest combination of 5 and 3 is 15, and 14 is very close to that, so this is an indication that it is likely that 5 and 3 must be multiplied together:

Set up pairs of factors and cross multiply them.
(Factors of 5 on the left and factors of 3 on the right)

Find the difference between the answers - just subtract the numbers, don't worry about the signs yet.

`color(white)(.x.)5color(white)(xxx.)3`
`color(white)(xx)darrcolor(white)(..x)darr`

`" " 5" "1 rarr 1 xx 1 = 1`
`" " 1" "3 rarr 5 xx3 = ul15`
`color(white)(xxxxxxxxxxxxx.xx)14" "larr` we have the correct factors!

Only now do we work out the signs...

We need `+14` (there are more positives)

`" " 5" "1 rarr 1 xx 1 = -1" " larr` must be `color(red)(- 1)`
`" " 1" "3 rarr 5 xx3 = ul(+15)" "larr` must be `color(blue)(+15)`
`color(white)(xxxxxxxxxxxxx.xx)+14" "larr (+15-1 = +14)`

Now insert the signs we have found with the factors:

the first bracket:`" " (5" "color(red)(-1) )rarr 1 xx color(red)(- 1= -1)`
second bracket:`" " (1" "color(blue)(+3)) rarr 5 xxcolor(blue)(+3 = ul(+15))`

Now we know exactly what the two brackets are:

`(5x-1)(x+3)`
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Note that the only other possible combination of factors would have been:

`" " 1" "1 rarr 5xx 1 = 5`
`" " 5" "3 rarr 1 xx3 = ul3`
`color(white)(xxxxxxxxxxxxx.xx)2" "larr` we have the wrong factors!
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