I need help with these math problems. It's about equations by factoring.?

Question

1. 12x^2+25x+12=0
2. 12x^2+18x+6=0
3. 7x^2-36x+5=0

Answer 1

1) 12x² + 25x +12=0
12x² + 16x + 9x +12 =0
4x(3x + 4) + 3(3x +4) = 0
(4x +3)(3x + 4) = 0
4x +3 = 0 or 3x + 4 = 0
x = -3/4 or x = -4/3

2) 12x²+18x+6=0
12x² + 12x + 6x + 6 =0
12x( x + 1) + 6(x+1) = 0
6(2x + 1)(x + 1) = 0
2x + 1 = 0 or x+ 1 = 0
x = -1/2 or x = -1

3) 7x² -36x +5=0
7x² - 35x - x + 5 =0
7x(x - 5) -1(x - 5) = 0
(7x - 5)(x - 5) = 0
x = 5/7 or x = 5

Answer 2

1. 12x^2 + 25x + 12 = 0
or 12x^2 + 16x + 9x + 12 = 0
or 4x*( 3x +4) + 3*(3x +4) = 0
or (4x +3) (3x +4 ) = 0
Therefore, x= -3/4 or -4/3

2. 12x^2 + 18x + 6 = 0
or 2x^2 + 3x + 2 = 0 (dividing both sides by 6)
or 2x^2 + 2x +x + 2 = 0
or 2x*(x+2) + 1*(x+2) = 0
or (2x+1) (x+2) = 0
x= -1/2 or -2

3. 7x^2 - 36x + 5 = 0
or 7x^2 - 35x - x + 5 = 0
or 7x*(x-5) -1 * (x-5) = 0
or (7x -1 ) (x-5 ) = 0
x = 1/7 or 5

Answer 3

I won't tell you all the answers instead I will tell you how to factorise the quadratics.

In a quadratic equation we have ax^2+bx+c so to factorise it we have to take the constant i.e c and multiply it by a. Then we have to find to factors of the resulting number which add up to make b. We then finally factorise the equation.

Let me show you an example, 12x^2+25x+12

a X c = 144

two facctors of 144 which add up to make 25 are 16 and 9 so we substitue them into the equation giving

12x^2+16x+9x+12

We then seperatly factorise 12x^2+16x and 9x+12
12x^2+16x gives 4x(3x+4)

9x+12 gives 3(3x+4)

So we get 4x(3x+4)+3(3x+4)

So we get (4x+3) (3x+4)

To solve it and find the value of x

4x= - 3

x= - 3/4

and 3x= - 4

x= - 4/3

Answer 4

I can see how these questions can seem a bit intimidating at first but......let's try to simplify it.......

1. You need to find the product of a A and C. So multiple in the case of eq. 1, 12*12=144.
2. Now you need to find two numbers that when multiplied equal 144, but when added, equal the middle coefficeint B.

So.... a+b=25, ab=144

3. You can fiddle with your calc, and find that if you divide 144 by 9, you get 16. 9+16= 25.

Now input 9 and 16 for the coefficent of C

12x^2+9x+16x+12=0

4. Factor out the common factors,

12x^2+9x have a common factor of 3x, so...

3x(4x+3)

and for 16x+12, they have a common factor of 4,

4(4x+3)

5. 3x(4x+3)+4(4X+3)

Take out the common factor (4x-3)

(4x-3)(3x-4)

Using the zero-product principle you'd set each one equal to zero and solve.

4x-3=0 3x-4=0
x=3/4 x=4/3

..........

Applying these steps to 2 you'll get.......

2. x=-1/2, -1


and I'm sure you can figure out three =)

Hoped this helped!

Answer 5

1: (4x+3)(3x+4) = 0 x1 = -.75 x2 = -4/3

2: (6x+3)(2x+2) = 0 x1 = -.50 x2 = -1

3: (7x-1)(x-5) = 0 x1 = 1/7 x2 = 5

Answer 6

1.(3x+4)(4x+3)=0 ,x=-4/3 or -3/4
2.6(2x+1)(x+1)=0 ,x=-1/2 or -1
3.(7x-1)(x-5)=0 ,x=1/7 or 5

Answer 7

1) (4x+3)(3x+4)

2) (6x+3)(2x+2)

3) (7x-1)(x-5)

Answer 8

(4x+3)(3x+4)=0

4x=-3
3x=-4
x=-4/3;-3/4

(6x+3)(3x+2)=0
6x=-3
3x=-2
x=-3/6=-1/2
x=-2/3

(7x-1)(x-5)=0
7x=1
x=1/7 x=5