Solve -3x^2+30x+33

$$- 3 x ^ { 2 } + 30 x + 33$$

Factor

$$-3\left(x-11\right)\left(x+1\right)$$

Solution Steps

Factor out 3.

$$3\left(-x^{2}+10x+11\right)$$

Consider -x^{2}+10x+11. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+11. To find a and b, set up a system to be solved.

$$a+b=10 ab=-11=-11$$

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. The only such pair is the system solution.

$$a=11 b=-1$$

Rewrite -x^{2}+10x+11 as \left(-x^{2}+11x\right)+\left(-x+11\right).

$$\left(-x^{2}+11x\right)+\left(-x+11\right)$$

Factor out -x in the first and -1 in the second group.

$$-x\left(x-11\right)-\left(x-11\right)$$

Factor out common term x-11 by using distributive property.

$$\left(x-11\right)\left(-x-1\right)$$

Rewrite the complete factored expression.

$$3\left(x-11\right)\left(-x-1\right)$$

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.

$$-3x^{2}+30x+33=0$$

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.

$$x=\frac{-30±\sqrt{30^{2}-4\left(-3\right)\times 33}}{2\left(-3\right)}$$

Square 30.

$$x=\frac{-30±\sqrt{900-4\left(-3\right)\times 33}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-30±\sqrt{900+12\times 33}}{2\left(-3\right)}$$

Multiply 12 times 33.

$$x=\frac{-30±\sqrt{900+396}}{2\left(-3\right)}$$

Add 900 to 396.

$$x=\frac{-30±\sqrt{1296}}{2\left(-3\right)}$$

Take the square root of 1296.

$$x=\frac{-30±36}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-30±36}{-6}$$

Now solve the equation x=\frac{-30±36}{-6} when ± is plus. Add -30 to 36.

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

Divide 6 by -6.

$$x=-1$$

Now solve the equation x=\frac{-30±36}{-6} when ± is minus. Subtract 36 from -30.

$$x=-\frac{66}{-6}$$

Divide -66 by -6.

$$x=11$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1 for x_{1} and 11 for x_{2}.

$$-3x^{2}+30x+33=-3\left(x-\left(-1\right)\right)\left(x-11\right)$$

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

$$-3x^{2}+30x+33=-3\left(x+1\right)\left(x-11\right)$$

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.

$$x ^ 2 -10x -11 = 0$$

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

$$r + s = 10 rs = -11$$

Two numbers r and s sum up to 10 exactly when the average of the two numbers is \frac{1}{2}*10 = 5. 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>

$$r = 5 - u s = 5 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -11

$$(5 - u) (5 + u) = -11$$

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

$$25 - u^2 = -11$$

Simplify the expression by subtracting 25 on both sides

$$-u^2 = -11-25 = -36$$

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

$$u^2 = 36 u = \pm\sqrt{36} = \pm 6$$

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

$$r =5 - 6 = -1 s = 5 + 6 = 11$$

Evaluate

$$-3\left(x-11\right)\left(x+1\right)$$