how do i factor these equations

Question

I seem to have a hard time factoring equations is there any trick to it

 

s²+36-12s

4r²-256

v³-49v²-4v+196

60s²-65s-175

c²+49

-5r⁴+10r³-25r

 

Answer 1

here's a couple for you...

 

v³-49v²-4v+196

(v³-49v²)+(-4v+196)        (I did this because 196/4=49)

v²(v-49)-4(v-49)

(v²-4)(v-49)

(v-2)(v+2)(v-49)

 

60s²-65s-175

5(12s²-13s-35)

12*35=2*2*3*5*7 and 2*2*7=28, 3*5=15, and 28-15=13, the coefficient of the middle term

5(4s+5)(3s-7)

 

c²+49 can't be factored using integer coefficients

c²+49=(c+7i)(c-7i) where i=√(-1)

if you meant c²-49 then you have the difference of two squares

 

-5r⁴+10r³-25r=-5r(r³-2r²+5)

can't seem to go any further with r³-2r²+5 without getting a remainder

Answer 2

First, get them in proper form -- the first equation, for instance, should be written s² - 12s + 36, and the last should be written r(-5r³ + 10r² + 0r - 25).   Then, use your norman factoring "tricks" -- what numbers added together would give you the coefficient of the middle number, which they multiplied together give you the final number.

 

For example, in Problem 1, -4 and -8 added together give you -12 and multiplied together give you +36, so the answer is (s - 4)(s - 8).

 

For Problem 2, factor out the 4, making it 4(r² - 64) . . . and, since 64 is the square of 8, the answer is 4(r - 8)(r - 8).

 

On Problems 3 and 6, they are trinomial factors, meaning in the form (r +/- x) (r +/- y) (r +/- z), with the final number being x times y times z.   Just keep trying combinations.....