Question

1. Solve by factoring
a. 3x^2-18x=0
b. x(3x-1)=2
c.(4a+1)(3a+1)=13
d.x^2-16x+48=0

Answer 1

a. \[3x ^{2}-18x=0\]
\[3x ^{2}=18\]
\[x ^{2}=9\]
\[x=3;x=-3\]

Answer 2

3m^3-m^2-18m=0

Answer 3

thank you guys for your help. i really appreciate it

Answer 4

\[x(3x-1)=2\]

\[3x ^{2}-x-2=0\]
\[(3x+2)(x-1)\]

Answer 5

c.\[(4a+1)(3a+1)=13\]
\[12a ^{2}+4a+3a+1-13=0\]
\[12a ^{2}+7a-12=0\]
\[(4a-3)(3a+4)\]

Answer 6

\[x ^{2}-16x+48=0\]
\[(x-12)(x-4)\]

Answer 7

ah yah @Euler271 is right sorry

Answer 8

the mistake in a is actually more than just that.

you don't consider when x = 0

you hypothetically divided by zero

Answer 9

proper way is 3x(x-6) = 0

x = 0 or x = 6

Answer 10

@abrown16 please consider @Euler271 answers for letter a mine is wrong :)

Answer 11

ok @yeyenunez thank you