Factor 3x² +7x+2?

Question

A.Factor 3x² + 7x +2

B. Find the solution set of 3x² = 4x

I am having trouble understanding how to perform these certain steps to solve these problems. please help me understand the problems. I dont want just answers. I want to have a good understanding of the subject. Thanks

Answer 1

3x^2 + 7x + 2 = 3(x^2 + 7/3 x +2/3)
= 3(x^2 + 7/3x + (7/6)^2 - (7/6)^2 +2/3)
= 3((x+ 7/6)^2 - (49/36 - 24/36))
= 3((x+7/6)^2 - 25/36)
= 3(x+7/6-5/6)(x+7/6+5/6)
=3(x + 1/3)(x + 2)


3x^2 - 4x = 0
x(3x - 4) = 0
x=0 and x=4/3

Answer 2

A. 3x^2 + 7x + 2
= 3x^2 +6x + x + 2
= 3x(x + 2) + 1(x + 2)
= (x + 2)(3x + 1) Answer. Take the product of the coefficient
of x^2 and the last term(constant). Find out two such factors
of that product which when added should be equal to the
coefficient of x. Then with those factors spread the middle
term and factorise. It is easy. But you practice one or two.

B 3x^2 = 4x
3x^2 - 4x = 0
x(3x - 4) = 0. So, either x = 0 or (3x - 4) = 0. If
3x - 4 = 0, 3x = 4. Therefore x = 4/3.
Therefore solution set = { 0, 4/3} Answer.

Answer 3

A. 3x^2 + 7x + 2 = 0 (you didn't put the zero, but I think you meant to).

First, you find all the multiple sequences of the 3 & 2 (A & C).
We have (1,3), (-1,-3), & (1,2,), (-1,-2). Using one set from each side, find a combination that can add up to 7. It's done like this
1*1 = 1
3*2 = 6
Do you see the sets I chose? (1,3) & (1,2). The answers are 1 & 6, which add up to 7. From here, we build our two parenthesis for factoring the original equation.
(3x + 1)(x + 2) (See hwo the first parenthesis contains 3,1, and the second contains 2,1. All we did was multiply an x to the first number of each parenthesis. When these two get multiplied out now, you'll have the same degree of x that the equation started with).

B. 3x^2 = 4x (subtract 4x from each side)

3x^2 - 4x = 0 (factor an x out)

x(3x - 4) = 0 (now we can break it apart)

x = 0 & (3x - 4) = 0 (first one is easy, x = 0. Now solve the second)

3x - 4 = 0 (add 4 to both sides)

3x = 4 (divide both sides by three)

x = 4/3

Answer 4

A.
1. First set the factor = 0.
2. Since you have no numbers that be reduced take out the the 3x.
3. You will have 2 factors. (3x+ _)(x + _)
4.The outer numbers when multiplied must equal 2 so use 2 and 1.
5. Since you have all positive numbers the factors will have a plus sign in between them
6. (3x + 1)(x + 2)

To find out if you have done it right use the FOIL method (FIRST)multiply 3x times x = 3x^2. (OUTER) Next 3x times 2 = 6x. (INNER)Then 1 times x = 1x (LAST) 1 times 2 = 2

B. 3x^2=4x

I saw people do it the other wasy so I decided to show you a different way. I would set it equal to 0 and then use the quadratic formula.

a=3 b= -4 c=0

minus b plus or minus (as in that a squared number could be positive or negative)sqaure root of b squared minus 4ac (all under the radical) divided by 2a.

Answer 5

A. (3x + 1)(x + 2)
Since this is all addition, just look at the factors of each coefficient (3, 1, 2). See how they multiply/add together and then test it out
(3x + 1)(x + 2) = 3x^2 + x + 6x + 2 = 3x^2 + 7x + 2

B. 3x^2 = 4x
x = 0
if not, divide both sides by x
3x = 4
x = 4/3
so x = 0, 4/3

Answer 6

A. Quadratic formula:
x ={ -7 +- sqr[7^2 - 4(3)(2)]}/2(3)
= {-7 +- sqsr[49-24]}/6 = {-7 +- 5}/6 = -2/6, -12/6 = -1/3, -2

B. 3x^2 = 4x divide both sides by 3x
x = 4/3

Answer 7

a. check all the possibilities
need facotrs of 3 and factors of 1 and the product of the factors need toadd to 7...since 3 has only factors of 3 and 1 and 1 is only 1 and 1
so (3x+1)(x+1)

b. set = 0
and get 3x^2-4x =0
check for GCF of terms (x)...and factor it out
x(3x-4)=0
set each = 0 and solve for x
x=0 and 3x-4 = 0
x=0 and x = 4/3

Answer 8

A. (3x+1) (x+2)
B. x = 0, 4/3

Answer 9

(3x+1)(x+2) is factored. Try to figure it out from here. REMEMBER FOIL!