Answer:
5(v -2)(v -4)
Explanation:
You want to factor completely 5v^2-30v+40.
Common constant
The coefficients in this quadratic have a common factor of 5. It works nicely to factor that out first.
= 5(v² -6v +8)
Product of factors
The product of two binomial factors looks like this:
(v -a)(v -b) = v² -(a+b)v +ab
Application
When you compare this to the quadratic factor (v² -6v +8), you see that ...
(a+b) = 6
ab = 8
To find values for 'a' and 'b', it usually works well to start by factoring the product:
ab = 8 = 1·8 = 2·4
The sums of these factor pairs are 1+8=9 and 2+4=6. The pair with a sum of 6 is what we're looking for. This means we can choose a=2, b=4 and our factorization is ...
5v² -30v +40 = 5(v -2)(v -4)
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Additional comment
There are some graphic aids that are sometimes used to help you find the 'a' and 'b' constants in the binomial factors. You may run across the "X" method, where the sum is put in the top (or bottom) of the X, and the product is put in the bottom (or top) of the X. Then the left and right sides are filled with the numbers having that sum and product.
That can help you focus on what you're doing, but the X fundamentally doesn't add anything to the solution process.
Quadratics with leading coefficient not 1
If the leading coefficient of the quadratic cannot be made to be 1, then a slight modification to the method shown here is needed. The factors of ax² +bx +c will be binomial terms generally of the form (px +q)(rx +s). If you play with all of the literals involved, you can convince yourself that you want to find factors of 'ac' that have a sum of 'b'. If those are 'u' and 'v', you can use them a couple of ways. One is to rewrite the quadratic as ...
ax² +ux +vx +c
which you can factor by grouping:
(ax² +ux) +(vx +c)
The other way to use these constants is to write the factorization as ...
(ax +u)(ax +v)/a
Then the division by 'a' can be used to remove constant factors from one or both binomials to reduce them to "prime" factors.