Solve 6u^2+19u+10

$$6 u ^ { 2 } + 19 u + 10$$

Factor

$$\left(2u+5\right)\left(3u+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6u^{2}+au+bu+10. To find a and b, set up a system to be solved.

$$a+b=19 ab=6\times 10=60$$

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 60.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Calculate the sum for each pair.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

The solution is the pair that gives sum 19.

$$a=4 b=15$$

Rewrite 6u^{2}+19u+10 as \left(6u^{2}+4u\right)+\left(15u+10\right).

$$\left(6u^{2}+4u\right)+\left(15u+10\right)$$

Factor out 2u in the first and 5 in the second group.

$$2u\left(3u+2\right)+5\left(3u+2\right)$$

Factor out common term 3u+2 by using distributive property.

$$\left(3u+2\right)\left(2u+5\right)$$

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.

$$6u^{2}+19u+10=0$$

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.

$$u=\frac{-19±\sqrt{19^{2}-4\times 6\times 10}}{2\times 6}$$

Square 19.

$$u=\frac{-19±\sqrt{361-4\times 6\times 10}}{2\times 6}$$

Multiply -4 times 6.

$$u=\frac{-19±\sqrt{361-24\times 10}}{2\times 6}$$

Multiply -24 times 10.

$$u=\frac{-19±\sqrt{361-240}}{2\times 6}$$

Add 361 to -240.

$$u=\frac{-19±\sqrt{121}}{2\times 6}$$

Take the square root of 121.

$$u=\frac{-19±11}{2\times 6}$$

Multiply 2 times 6.

$$u=\frac{-19±11}{12}$$

Now solve the equation u=\frac{-19±11}{12} when ± is plus. Add -19 to 11.

$$u=-\frac{8}{12}$$

Reduce the fraction \frac{-8}{12} to lowest terms by extracting and canceling out 4.

$$u=-\frac{2}{3}$$

Now solve the equation u=\frac{-19±11}{12} when ± is minus. Subtract 11 from -19.

$$u=-\frac{30}{12}$$

Reduce the fraction \frac{-30}{12} to lowest terms by extracting and canceling out 6.

$$u=-\frac{5}{2}$$

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

$$6u^{2}+19u+10=6\left(u-\left(-\frac{2}{3}\right)\right)\left(u-\left(-\frac{5}{2}\right)\right)$$

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

$$6u^{2}+19u+10=6\left(u+\frac{2}{3}\right)\left(u+\frac{5}{2}\right)$$

Add \frac{2}{3} to u by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6u^{2}+19u+10=6\times \frac{3u+2}{3}\left(u+\frac{5}{2}\right)$$

Add \frac{5}{2} to u by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6u^{2}+19u+10=6\times \frac{3u+2}{3}\times \frac{2u+5}{2}$$

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

$$6u^{2}+19u+10=6\times \frac{\left(3u+2\right)\left(2u+5\right)}{3\times 2}$$

Multiply 3 times 2.

$$6u^{2}+19u+10=6\times \frac{\left(3u+2\right)\left(2u+5\right)}{6}$$

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

$$6u^{2}+19u+10=\left(3u+2\right)\left(2u+5\right)$$

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

$$x ^ 2 +\frac{19}{6}x +\frac{5}{3} = 0$$

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

$$r + s = -\frac{19}{6} rs = \frac{5}{3}$$

Two numbers r and s sum up to -\frac{19}{6} exactly when the average of the two numbers is \frac{1}{2}*-\frac{19}{6} = -\frac{19}{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>

$$r = -\frac{19}{12} - u s = -\frac{19}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{3}

$$(-\frac{19}{12} - u) (-\frac{19}{12} + u) = \frac{5}{3}$$

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

$$\frac{361}{144} - u^2 = \frac{5}{3}$$

Simplify the expression by subtracting \frac{361}{144} on both sides

$$-u^2 = \frac{5}{3}-\frac{361}{144} = -\frac{121}{144}$$

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.

$$r =-\frac{19}{12} - \frac{11}{12} = -2.500 s = -\frac{19}{12} + \frac{11}{12} = -0.667$$

Evaluate

$$\left(2u+5\right)\left(3u+2\right)$$