Solve 6n^2+31n+35

$$6 n ^ { 2 } + 31 n + 35$$

Factor

$$\left(2n+7\right)\left(3n+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6n^{2}+an+bn+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 positive, a and b are both positive. 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.

$$a=10 b=21$$

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

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

Factor out 2n in the first and 7 in the second group.

$$2n\left(3n+5\right)+7\left(3n+5\right)$$

Factor out common term 3n+5 by using distributive property.

$$\left(3n+5\right)\left(2n+7\right)$$

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.

$$6n^{2}+31n+35=0$$

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.

$$n=\frac{-31±\sqrt{31^{2}-4\times 6\times 35}}{2\times 6}$$

Square 31.

$$n=\frac{-31±\sqrt{961-4\times 6\times 35}}{2\times 6}$$

Multiply -4 times 6.

$$n=\frac{-31±\sqrt{961-24\times 35}}{2\times 6}$$

Multiply -24 times 35.

$$n=\frac{-31±\sqrt{961-840}}{2\times 6}$$

Add 961 to -840.

$$n=\frac{-31±\sqrt{121}}{2\times 6}$$

Take the square root of 121.

$$n=\frac{-31±11}{2\times 6}$$

Multiply 2 times 6.

$$n=\frac{-31±11}{12}$$

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

$$n=-\frac{20}{12}$$

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

$$n=-\frac{5}{3}$$

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

$$n=-\frac{42}{12}$$

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

$$n=-\frac{7}{2}$$

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

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

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

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

Add \frac{5}{3} to n by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Add \frac{7}{2} to n by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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

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

Multiply 3 times 2.

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

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

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

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.

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

Evaluate

$$\left(2n+7\right)\left(3n+5\right)$$