Solve -3x^2-8x+3

$$- 3 x ^ { 2 } - 8 x + 3$$

Factor

$$-\left(3x-1\right)\left(x+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -3x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-8 ab=-3\times 3=-9$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -9.

$$1,-9 3,-3$$

Calculate the sum for each pair.

$$1-9=-8 3-3=0$$

The solution is the pair that gives sum -8.

$$a=1 b=-9$$

Rewrite -3x^{2}-8x+3 as \left(-3x^{2}+x\right)+\left(-9x+3\right).

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

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

$$-x\left(3x-1\right)-3\left(3x-1\right)$$

Factor out common term 3x-1 by using distributive property.

$$\left(3x-1\right)\left(-x-3\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-3x^{2}-8x+3=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-3\right)\times 3}}{2\left(-3\right)}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-3\right)\times 3}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-8\right)±\sqrt{64+12\times 3}}{2\left(-3\right)}$$

Multiply 12 times 3.

$$x=\frac{-\left(-8\right)±\sqrt{64+36}}{2\left(-3\right)}$$

Add 64 to 36.

$$x=\frac{-\left(-8\right)±\sqrt{100}}{2\left(-3\right)}$$

Take the square root of 100.

$$x=\frac{-\left(-8\right)±10}{2\left(-3\right)}$$

The opposite of -8 is 8.

$$x=\frac{8±10}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{8±10}{-6}$$

Now solve the equation x=\frac{8±10}{-6} when ± is plus. Add 8 to 10.

$$x=\frac{18}{-6}$$

Divide 18 by -6.

$$x=-3$$

Now solve the equation x=\frac{8±10}{-6} when ± is minus. Subtract 10 from 8.

$$x=-\frac{2}{-6}$$

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

$$x=\frac{1}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -3 for x_{1} and \frac{1}{3} for x_{2}.

$$-3x^{2}-8x+3=-3\left(x-\left(-3\right)\right)\left(x-\frac{1}{3}\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$-3x^{2}-8x+3=-3\left(x+3\right)\left(x-\frac{1}{3}\right)$$

Subtract \frac{1}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-3x^{2}-8x+3=-3\left(x+3\right)\times \frac{-3x+1}{-3}$$

Cancel out 3, the greatest common factor in -3 and 3.

$$-3x^{2}-8x+3=\left(x+3\right)\left(-3x+1\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 +\frac{8}{3}x -1 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -\frac{8}{3} rs = -1$$

Two numbers r and s sum up to -\frac{8}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{8}{3} = -\frac{4}{3}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{4}{3} - u s = -\frac{4}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -1

$$(-\frac{4}{3} - u) (-\frac{4}{3} + u) = -1$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{16}{9} - u^2 = -1$$

Simplify the expression by subtracting \frac{16}{9} on both sides

$$-u^2 = -1-\frac{16}{9} = -\frac{25}{9}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{25}{9} u = \pm\sqrt{\frac{25}{9}} = \pm \frac{5}{3}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-\frac{4}{3} - \frac{5}{3} = -3 s = -\frac{4}{3} + \frac{5}{3} = 0.333$$

Evaluate

$$-\left(3x-1\right)\left(x+3\right)$$