# I need help solve for x x(x-4)(3x+2)=240

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

x(x-4)(3x+2) = 240

First we will open brackets:

(x^2 - 4x)(3x+2) = 240

==> 3x^3 + 2x^2 -12x^2 - 8x - 240 = 0

==> 3x^3 -10x^2 - 8x - 240 =0

Now we will try and substitute with 240 factors.

==> x= 1 ==> 3-10-8 -240 = -255

==> x= 3 ==> 27 - 90 - 8 - 240 = -311

==> x= 6 ==> 648 - 360 - 48 - 240 = 0

Then we find that x= 6 is one of the roots ... Then (x-6) is a facto.

Now we will divide the equation by (x-6) to find the other factors.

==> (x-6)(3x^2 +8x + 40) = 0

==> x1= 6

Now we will use the quadratic equation to find the other 2 roots.

==> x2= ( -8 + sqrt( 64-4*3*40)/ 2*3 = ( -8 + sqrt(416)*i)/6= -8+4sqrt26 )/ 6 = (4/3) + (2sqrt26/3)*i

==> x2= (4/3) - (2sqrt26/3)*i

Then the roots are: x= { 6, (-4/3)+(2sqrt26/3)*i  , (-4/3 - (sqrt26/3)*i}

givingiswinning | Student, Grade 10 | (Level 1) Valedictorian

Posted on

x(x-4)(3x+2)=240

x(x-4)(3x+2)

(x^2 - 4x)(3x+2) = 240

3x^3 + 2x^2 -12x^2 - 8x =  240

3x^3 -10x^2 - 8x - 240

atyourservice | Student, Grade 11 | (Level 3) Valedictorian

Posted on

x(x-4)(3x+2)=240

x(x-4) l (3x+2)

distribute the x to the numbers in the first parentheses:

(x^2 - 4x) l (3x+2) = 240

now foil the next 2 parentheses:

3x^3 + 2x^2 -12x^2 - 8x =  240

combine like terms:

3x^3 -10x^2 - 8x - 240

find factors of 240 and plug them in to see which number would make the equations equal 0. which would be 6 so x would be 6. Since you already have 1 root, use the quadratic formula to get the other 2.