18x^2+21x-9 I need to figure out how to work this problem

2 answers:

3 0

$18x^2+21x-9=3(6x^2)+3(7x)-3(3)\\\\=3(6x^2+7x-3)=3(6x^2+9x-2x-3)\\\\=3[3x(2x)+3x(3)-1(2x)-1(3)]\\\\=3[3x(2x+3)-1(2x+3)]\\\\=\boxed{3(2x+3)(3x-1)}$

Goshia [24]3 years ago

3 0

18x^2 + 21x - 9. Factor out a 3.

3(6x^2 + 7x - 3). Factor the trinomial. 

3(3x - 1)(2x + 3). Answer

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Solve the equation 3/4x+3-2x = -1/4+1/2x+5

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Hi there! The answer is x = -1

Let's solve this equation step by step!
$\frac{3}{4} x + 3 - 2x = - \frac{1}{4} + \frac{1}{2} x + 5$

First collect terms.
$- 1 \frac{1}{4} x + 3 = \frac{1}{2} x + 4 \frac{3}{4}$

Subtract (1/2)x from both sides.
$- 1\frac{3}{4} x + 3 = 4 \frac{3}{4}$

Subtract 3 from both sides.
$- 1\frac{3}{4} x = 1 \frac{3}{4}$

And finally divide by -1 3/4
$x = \frac{7}{4} \div - \frac{7}{4} = \frac{7}{4} \times - \frac{4}{7} = - \frac{28}{28} = - 1$