Solve 3x^2+33x+30

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

Factor

$$3\left(x+1\right)\left(x+10\right)$$

Solution Steps

Factor out 3.

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

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

$$a+b=11 ab=1\times 10=10$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 10.

$$1,10 2,5$$

Calculate the sum for each pair.

$$1+10=11 2+5=7$$

The solution is the pair that gives sum 11.

$$a=1 b=10$$

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

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

Factor out x in the first and 10 in the second group.

$$x\left(x+1\right)+10\left(x+1\right)$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(x+10\right)$$

Rewrite the complete factored expression.

$$3\left(x+1\right)\left(x+10\right)$$

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.

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

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.

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

Square 33.

$$x=\frac{-33±\sqrt{1089-4\times 3\times 30}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-33±\sqrt{1089-12\times 30}}{2\times 3}$$

Multiply -12 times 30.

$$x=\frac{-33±\sqrt{1089-360}}{2\times 3}$$

Add 1089 to -360.

$$x=\frac{-33±\sqrt{729}}{2\times 3}$$

Take the square root of 729.

$$x=\frac{-33±27}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-33±27}{6}$$

Now solve the equation x=\frac{-33±27}{6} when ± is plus. Add -33 to 27.

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

Divide -6 by 6.

$$x=-1$$

Now solve the equation x=\frac{-33±27}{6} when ± is minus. Subtract 27 from -33.

$$x=-\frac{60}{6}$$

Divide -60 by 6.

$$x=-10$$

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 -10 for x_{2}.

$$3x^{2}+33x+30=3\left(x-\left(-1\right)\right)\left(x-\left(-10\right)\right)$$

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

$$3x^{2}+33x+30=3\left(x+1\right)\left(x+10\right)$$

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.This is achieved by dividing both sides of the equation by 3

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

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

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

Two numbers r and s sum up to -11 exactly when the average of the two numbers is \frac{1}{2}*-11 = -\frac{11}{2}. 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>

$$r = -\frac{11}{2} - u s = -\frac{11}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 10

$$(-\frac{11}{2} - u) (-\frac{11}{2} + u) = 10$$

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

$$\frac{121}{4} - u^2 = 10$$

Simplify the expression by subtracting \frac{121}{4} on both sides

$$-u^2 = 10-\frac{121}{4} = -\frac{81}{4}$$

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

$$u^2 = \frac{81}{4} u = \pm\sqrt{\frac{81}{4}} = \pm \frac{9}{2}$$

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

$$r =-\frac{11}{2} - \frac{9}{2} = -10 s = -\frac{11}{2} + \frac{9}{2} = -1$$

Evaluate

$$3\left(x+1\right)\left(x+10\right)$$