Solve (4x-3)(x^2+x+3)=(2x+1)(2x^2-3x)

$$( 4 x - 3 ) ( x ^ { 2 } + x + 3 ) = ( 2 x + 1 ) ( 2 x ^ { 2 } - 3 x )$$

Solve for x

$$x=-3$$

$$x=\frac{3}{5}=0.6$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 4x-3 by x^{2}+x+3 and combine like terms.

$$4x^{3}+x^{2}+9x-9=\left(2x+1\right)\left(2x^{2}-3x\right)$$

Use the distributive property to multiply 2x+1 by 2x^{2}-3x and combine like terms.

$$4x^{3}+x^{2}+9x-9=4x^{3}-4x^{2}-3x$$

Subtract 4x^{3} from both sides.

$$4x^{3}+x^{2}+9x-9-4x^{3}=-4x^{2}-3x$$

Combine 4x^{3} and -4x^{3} to get 0.

$$x^{2}+9x-9=-4x^{2}-3x$$

Add 4x^{2} to both sides.

$$x^{2}+9x-9+4x^{2}=-3x$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}+9x-9=-3x$$

Add 3x to both sides.

$$5x^{2}+9x-9+3x=0$$

Combine 9x and 3x to get 12x.

$$5x^{2}+12x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=12 ab=5\left(-9\right)=-45$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -45.

$$-1,45 -3,15 -5,9$$

Calculate the sum for each pair.

$$-1+45=44 -3+15=12 -5+9=4$$

The solution is the pair that gives sum 12.

$$a=-3 b=15$$

Rewrite 5x^{2}+12x-9 as \left(5x^{2}-3x\right)+\left(15x-9\right).

$$\left(5x^{2}-3x\right)+\left(15x-9\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(5x-3\right)+3\left(5x-3\right)$$

Factor out common term 5x-3 by using distributive property.

$$\left(5x-3\right)\left(x+3\right)$$

To find equation solutions, solve 5x-3=0 and x+3=0.

$$x=\frac{3}{5} x=-3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4x-3 by x^{2}+x+3 and combine like terms.

$$4x^{3}+x^{2}+9x-9=\left(2x+1\right)\left(2x^{2}-3x\right)$$

Use the distributive property to multiply 2x+1 by 2x^{2}-3x and combine like terms.

$$4x^{3}+x^{2}+9x-9=4x^{3}-4x^{2}-3x$$

Subtract 4x^{3} from both sides.

$$4x^{3}+x^{2}+9x-9-4x^{3}=-4x^{2}-3x$$

Combine 4x^{3} and -4x^{3} to get 0.

$$x^{2}+9x-9=-4x^{2}-3x$$

Add 4x^{2} to both sides.

$$x^{2}+9x-9+4x^{2}=-3x$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}+9x-9=-3x$$

Add 3x to both sides.

$$5x^{2}+9x-9+3x=0$$

Combine 9x and 3x to get 12x.

$$5x^{2}+12x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 12 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 5\left(-9\right)}}{2\times 5}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 5\left(-9\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-12±\sqrt{144-20\left(-9\right)}}{2\times 5}$$

Multiply -20 times -9.

$$x=\frac{-12±\sqrt{144+180}}{2\times 5}$$

Add 144 to 180.

$$x=\frac{-12±\sqrt{324}}{2\times 5}$$

Take the square root of 324.

$$x=\frac{-12±18}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-12±18}{10}$$

Now solve the equation x=\frac{-12±18}{10} when ± is plus. Add -12 to 18.

$$x=\frac{6}{10}$$

Reduce the fraction \frac{6}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{3}{5}$$

Now solve the equation x=\frac{-12±18}{10} when ± is minus. Subtract 18 from -12.

$$x=-\frac{30}{10}$$

Divide -30 by 10.

$$x=-3$$

The equation is now solved.

$$x=\frac{3}{5} x=-3$$

Steps for Completing the Square

Use the distributive property to multiply 4x-3 by x^{2}+x+3 and combine like terms.

$$4x^{3}+x^{2}+9x-9=\left(2x+1\right)\left(2x^{2}-3x\right)$$

Use the distributive property to multiply 2x+1 by 2x^{2}-3x and combine like terms.

$$4x^{3}+x^{2}+9x-9=4x^{3}-4x^{2}-3x$$

Subtract 4x^{3} from both sides.

$$4x^{3}+x^{2}+9x-9-4x^{3}=-4x^{2}-3x$$

Combine 4x^{3} and -4x^{3} to get 0.

$$x^{2}+9x-9=-4x^{2}-3x$$

Add 4x^{2} to both sides.

$$x^{2}+9x-9+4x^{2}=-3x$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}+9x-9=-3x$$

Add 3x to both sides.

$$5x^{2}+9x-9+3x=0$$

Combine 9x and 3x to get 12x.

$$5x^{2}+12x-9=0$$

Add 9 to both sides. Anything plus zero gives itself.

$$5x^{2}+12x=9$$

Divide both sides by 5.

$$\frac{5x^{2}+12x}{5}=\frac{9}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{12}{5}x=\frac{9}{5}$$

Divide \frac{12}{5}, the coefficient of the x term, by 2 to get \frac{6}{5}. Then add the square of \frac{6}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{12}{5}x+\left(\frac{6}{5}\right)^{2}=\frac{9}{5}+\left(\frac{6}{5}\right)^{2}$$

Square \frac{6}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{12}{5}x+\frac{36}{25}=\frac{9}{5}+\frac{36}{25}$$

Add \frac{9}{5} to \frac{36}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{12}{5}x+\frac{36}{25}=\frac{81}{25}$$

Factor x^{2}+\frac{12}{5}x+\frac{36}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{6}{5}\right)^{2}=\frac{81}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{6}{5}\right)^{2}}=\sqrt{\frac{81}{25}}$$

Simplify.

$$x+\frac{6}{5}=\frac{9}{5} x+\frac{6}{5}=-\frac{9}{5}$$

Subtract \frac{6}{5} from both sides of the equation.

$$x=\frac{3}{5} x=-3$$