2x^2= 7x - 3............... Factoring?

Question

please explain how to do this problem.

i know......

2x^2 - 7x +3 = 0

right??

then i don't know the next step, please help and explain how you do it. thank you

Answer 1

Start with the following....

(2x +  A)(x + B )

Using FOIL (first, outside, inside, last) for multiplication as your guide,

try to think of two numbers such that

2*B + A = -7

and

A*B = 3

The second part is easy, since 3 is prime, and positive, the options are

(1 and 3) or (-1 and -3)

so try

(2x - 1)(x - 3)

Multiply using FOIL

2x*x + (-1)*x + 2x*(-3) + (-1)(-3)

2x^2 - 7x + 3

So your answer is

(2x - 1)(x - 3) = 0

With practice, you'll get faster.

Answer 2

correct.  The next step is to factor it into a format of (ax+b)(cx+d), where

ac = 2

ac +bd = -7

cd = 3

(2x-1) (x-3)

Answer 3

(2x-1)(x-3)

Answer 4

When you factor you inow your formula will be of the form

( Ax   +  b)(Cx + d)= 0

from you equation you see that A*C=2. the only integers that will satis that are 2 and 1 by substituting for A and C

(2x + b)(1x + d)=0

now to find b & d: b*d=3 so either b=1(plus or minus) and d=3(Plus or minus) or vice versa

we know that b + 2d = -7

d has to be either -3 or -1 and to satisfy this equation, d can only = -3

so b has to be -1

(2x - 1)(x - 3)=0