Solve 6x^2-31x+35

$$6 x ^ { 2 } - 31 x + 35$$

Factor

$$\left(2x-7\right)\left(3x-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx+35. To find a and b, set up a system to be solved.

$$a+b=-31 ab=6\times 35=210$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 210.

$$-1,-210 -2,-105 -3,-70 -5,-42 -6,-35 -7,-30 -10,-21 -14,-15$$

Calculate the sum for each pair.

$$-1-210=-211 -2-105=-107 -3-70=-73 -5-42=-47 -6-35=-41 -7-30=-37 -10-21=-31 -14-15=-29$$

The solution is the pair that gives sum -31.

$$a=-21 b=-10$$

Rewrite 6x^{2}-31x+35 as \left(6x^{2}-21x\right)+\left(-10x+35\right).

$$\left(6x^{2}-21x\right)+\left(-10x+35\right)$$

Factor out 3x in the first and -5 in the second group.

$$3x\left(2x-7\right)-5\left(2x-7\right)$$

Factor out common term 2x-7 by using distributive property.

$$\left(2x-7\right)\left(3x-5\right)$$

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.

$$6x^{2}-31x+35=0$$

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.

$$x=\frac{-\left(-31\right)±\sqrt{\left(-31\right)^{2}-4\times 6\times 35}}{2\times 6}$$

Square -31.

$$x=\frac{-\left(-31\right)±\sqrt{961-4\times 6\times 35}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-31\right)±\sqrt{961-24\times 35}}{2\times 6}$$

Multiply -24 times 35.

$$x=\frac{-\left(-31\right)±\sqrt{961-840}}{2\times 6}$$

Add 961 to -840.

$$x=\frac{-\left(-31\right)±\sqrt{121}}{2\times 6}$$

Take the square root of 121.

$$x=\frac{-\left(-31\right)±11}{2\times 6}$$

The opposite of -31 is 31.

$$x=\frac{31±11}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{31±11}{12}$$

Now solve the equation x=\frac{31±11}{12} when ± is plus. Add 31 to 11.

$$x=\frac{42}{12}$$

Reduce the fraction \frac{42}{12} to lowest terms by extracting and canceling out 6.

$$x=\frac{7}{2}$$

Now solve the equation x=\frac{31±11}{12} when ± is minus. Subtract 11 from 31.

$$x=\frac{20}{12}$$

Reduce the fraction \frac{20}{12} to lowest terms by extracting and canceling out 4.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7}{2} for x_{1} and \frac{5}{3} for x_{2}.

$$6x^{2}-31x+35=6\left(x-\frac{7}{2}\right)\left(x-\frac{5}{3}\right)$$

Subtract \frac{7}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-31x+35=6\times \frac{2x-7}{2}\left(x-\frac{5}{3}\right)$$

Subtract \frac{5}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-31x+35=6\times \frac{2x-7}{2}\times \frac{3x-5}{3}$$

Multiply \frac{2x-7}{2} times \frac{3x-5}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-31x+35=6\times \frac{\left(2x-7\right)\left(3x-5\right)}{2\times 3}$$

Multiply 2 times 3.

$$6x^{2}-31x+35=6\times \frac{\left(2x-7\right)\left(3x-5\right)}{6}$$

Cancel out 6, the greatest common factor in 6 and 6.

$$6x^{2}-31x+35=\left(2x-7\right)\left(3x-5\right)$$

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 6

$$x ^ 2 -\frac{31}{6}x +\frac{35}{6} = 0$$

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

$$r + s = \frac{31}{6} rs = \frac{35}{6}$$

Two numbers r and s sum up to \frac{31}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{31}{6} = \frac{31}{12}. 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>

$$r = \frac{31}{12} - u s = \frac{31}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{35}{6}

$$(\frac{31}{12} - u) (\frac{31}{12} + u) = \frac{35}{6}$$

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

$$\frac{961}{144} - u^2 = \frac{35}{6}$$

Simplify the expression by subtracting \frac{961}{144} on both sides

$$-u^2 = \frac{35}{6}-\frac{961}{144} = -\frac{121}{144}$$

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

$$u^2 = \frac{121}{144} u = \pm\sqrt{\frac{121}{144}} = \pm \frac{11}{12}$$

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

$$r =\frac{31}{12} - \frac{11}{12} = 1.667 s = \frac{31}{12} + \frac{11}{12} = 3.500$$

Evaluate

$$\left(2x-7\right)\left(3x-5\right)$$