Question

How do you factor `8v^2-30v+18`?

Answer 1

2(4v - 3)(v - 3)

Explanation:

Use the systematic new AC Method (Socratic Search).
`f(v) = 8v^2 - 30v + 18 =` 8(v + p)(v + q)
Converted trinomial: `f'(v) = v^2 - 30v + 144 =` (v + p')(v + q').
p' and q' have same sign because ac > 0.
Factor pairs of (ac = 144) ---> (4, 36)(-4, -36)(6, 24)(-6, -24)..This sum is (-30 = b). Then p' = -6 and q' = -24.
Back to original f(v), `p = (p')/a = -6/8 = -3/4`, and `q = (q')/a = -24/8 = -3`.
Factored form: `f(v) = 8(v - 3/4)(v - 3) = 2(4v - 3)(v - 3).`

Answer 2

Solution reworked

Explanation:

Given;`" "8v^2-30v+18`

Sometimes you can spot the factors that work easily. Other times it takes a little work.

Notice that the coefficients and constant are even. So we can factor 2 out giving:

`" "2(4v^2-15v+9)`

So now we see if we can factorise `(4v^2-15+9)`
Consider the whole number factors of 4`->`{1,4} ; {2,2}
Consider the whole number factors of 9`->`{1,9} ; {3,3}
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Notice that `(3xx4)+(3xx1)=12+4=15`
But we need `-15` so we need to configure this multiplication to give as that value.

Try:`" "(4v-3)(v-3)` this gives:`" "4v^2-12v-3v+9` This works
;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So the full factorisation is:

`color(blue)(" "2(4v-3)(v-3))`

`color(red)("If this factorisation is correct it should have the same graph curve as the original equation."`