how do I factor this equation?? [alg. 1]?

i just use multiples of the first and last term and try to add/subtract them so that they become the second term

the answer is (5c-1)(c+1)

the multiples i chose for the first term were 5 and 1
the multiples i chose for the last term were 1 and 1

and 5-1 = 4

make
5c^2+4c-1=5[(c^2+4c/5-1/5]=0
solve it for c
c(1)=1/5
c(2)=-1
5c^2+4c-1=5(c+1)(c-1/5)
that's all

this is how u do it
the term that comes before c^2 is 5 (in this case)
the constant term is 1 (ignore the negative sign)

now if u add or subtract these terms u should get the
term before c in the eqn(in this case 4)

in this case its subtraction : 5-1 =4

so the eqn becomes

5c^2+5c-1c -1 (look at the middle part 4c=5c-1c)
now see "5c" is common in first two terms
and "-1" in the last two

take them out
5c(c+1) -1(c+1)

now (c+1) is common to "5c" and "-1"
take it out

(c+1)(5c-1)

this becomes complex when u have bigger numbers involved

team jointly the 7 and 3. it quite is because of the fact interior the equation [7(x+4)+3(x+4)], (x+4) is the consumer-friendly ingredient when you consider that 7(x+4) and 3(x+4) are each and every a term. so which you're doing away with the (x+4) and you're left with the 7 and 3. consequently: (7+3)(x+4) simplifies on your very final answer: 10(x+4). =]

(5c - 1)(c + 1)

5c^2+4c-1.
Take the product of coefficient of c^2 and the constant.
5*-1=-5
Now, you should find the factors of -5 such that their sum is 4. They are none other than 5 and -1.
Now split 4c as 5c and -c.
So, 5c^2+4c-1
=5c^2+5c-c-1
=(5c^2+5c)+(-c-1)
=5c(c+1)-1(c+1)
=(c+1)(5c-1)

5c^2 + 4c - 1
= (5c - 1)(c + 1)

= 5c^2 + 5c -1c -1

= 5c( c +1) - 1( c+1)


= (5c -1) (c+1)

(c+1)*(5*c-1)

(5c-1)(c+1)