How do you solve a trinomial?

Question

Please show me in detail how to factor the following trinomial completely.

5c^2-30c+40

Note : 5c(to the 2nd power)-30c+40

Answer 1

5c² - 30c + 40 =
first factor out the common 5
5(c² - 6c + 8)
now look for factors of 8 that add up to 6 (both negative)
5(c - 4)(c - 2)

Answer 2

first, common factor by taking out the 5
5c^2 - 30c + 40
= 5 (c^2 - 6c + 8) -> then use the grouping technique
= 5 (c^2 - 2c - 4c + 8) -> break -6c into two parts so that when multiplied, it equals 8, and when added, it equals -6c
= 5 (c^2 - 2c) - (4c - 8) -> put into 2 brackets (remember to change the signs for the 2nd bracket!)
= 5 [c(c - 2) - 4(c - 2)] -> common factor the c and the 4 out
= 5 (c - 2) (c-4) -> common factor the (c-2)

When you set the equation to 0...
5 (c-2) (c-4) = 0
Then c-2 or c-4 must equal to zero for the entire equation to equal 0
so...
5 (c-2) (c-4) = 0
Therefore,
c-2 = 0 OR c-4 = 0
c = 2 OR c = 4

These are your two answers =)

Answer 3

First, factor a 5 out of the equation, leaving you with 5(c^2-6c+8). Next, factor the reduced trinomial. (c^2-6c+8) = (c-2)(c-4). This factorization is typically trial and error for beginning students, and even still the case on occasion at the college level. Replacing c^2-6c+8 with its factorization gives the complete factorization of 5(c-2)(c-4)

Answer 4

the best way to solve most trinomials

5c^2 - 30c + 40
5(c^2 - 6c + 8)
5(c - 4)(c - 2)

c = 4 or 2

Answer 5

you get your equation equal to zero, then you can use the quadratic formula to find the x-values which give zeroes.
the quadratic equation will be in the general form
ax^2 + bx + c.
so using the quadratic formula, which is
x= [-b (plus or minus) sqrt (b^2 - 4ac) ] / 2a
so , in your case, it would be
x = [30 (plus or minus) sqrt (900-800) ] / 10
which gives you 30 +/- 10 divided by 10
which equals 20/10, or 2
and 40/10, or 4
so your x's are 4 and 2.
then you put those back into the original equation to find the y-values of the x-intercepts.

Answer 6

What Phil said.