Solve 3x^2+31x-34

$$3 x ^ { 2 } + 31 x - 34$$

Factor

$$\left(x-1\right)\left(3x+34\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-34. To find a and b, set up a system to be solved.

$$a+b=31 ab=3\left(-34\right)=-102$$

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 -102.

$$-1,102 -2,51 -3,34 -6,17$$

Calculate the sum for each pair.

$$-1+102=101 -2+51=49 -3+34=31 -6+17=11$$

The solution is the pair that gives sum 31.

$$a=-3 b=34$$

Rewrite 3x^{2}+31x-34 as \left(3x^{2}-3x\right)+\left(34x-34\right).

$$\left(3x^{2}-3x\right)+\left(34x-34\right)$$

Factor out 3x in the first and 34 in the second group.

$$3x\left(x-1\right)+34\left(x-1\right)$$

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

$$\left(x-1\right)\left(3x+34\right)$$

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.

$$3x^{2}+31x-34=0$$

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.

$$x=\frac{-31±\sqrt{31^{2}-4\times 3\left(-34\right)}}{2\times 3}$$

Square 31.

$$x=\frac{-31±\sqrt{961-4\times 3\left(-34\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-31±\sqrt{961-12\left(-34\right)}}{2\times 3}$$

Multiply -12 times -34.

$$x=\frac{-31±\sqrt{961+408}}{2\times 3}$$

Add 961 to 408.

$$x=\frac{-31±\sqrt{1369}}{2\times 3}$$

Take the square root of 1369.

$$x=\frac{-31±37}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-31±37}{6}$$

Now solve the equation x=\frac{-31±37}{6} when ± is plus. Add -31 to 37.

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

Divide 6 by 6.

$$x=1$$

Now solve the equation x=\frac{-31±37}{6} when ± is minus. Subtract 37 from -31.

$$x=-\frac{68}{6}$$

Reduce the fraction \frac{-68}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{34}{3}$$

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 -\frac{34}{3} for x_{2}.

$$3x^{2}+31x-34=3\left(x-1\right)\left(x-\left(-\frac{34}{3}\right)\right)$$

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

$$3x^{2}+31x-34=3\left(x-1\right)\left(x+\frac{34}{3}\right)$$

Add \frac{34}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3x^{2}+31x-34=3\left(x-1\right)\times \frac{3x+34}{3}$$

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

$$3x^{2}+31x-34=\left(x-1\right)\left(3x+34\right)$$

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

$$x ^ 2 +\frac{31}{3}x -\frac{34}{3} = 0$$

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

$$r + s = -\frac{31}{3} rs = -\frac{34}{3}$$

Two numbers r and s sum up to -\frac{31}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{31}{3} = -\frac{31}{6}. 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>

$$r = -\frac{31}{6} - u s = -\frac{31}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{34}{3}

$$(-\frac{31}{6} - u) (-\frac{31}{6} + u) = -\frac{34}{3}$$

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

$$\frac{961}{36} - u^2 = -\frac{34}{3}$$

Simplify the expression by subtracting \frac{961}{36} on both sides

$$-u^2 = -\frac{34}{3}-\frac{961}{36} = -\frac{1369}{36}$$

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

$$u^2 = \frac{1369}{36} u = \pm\sqrt{\frac{1369}{36}} = \pm \frac{37}{6}$$

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

$$r =-\frac{31}{6} - \frac{37}{6} = -11.333 s = -\frac{31}{6} + \frac{37}{6} = 1$$

Evaluate

$$\left(x-1\right)\left(3x+34\right)$$