Solve 10x^2+19x+6

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

Factor

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

Solution Steps

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

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

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

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

Factor out 2x in the first and 3 in the second group.

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

Factor out common term 5x+2 by using distributive property.

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

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.

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

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.

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

Square 19.

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

Multiply -4 times 10.

$$x=\frac{-19±\sqrt{361-40\times 6}}{2\times 10}$$

Multiply -40 times 6.

$$x=\frac{-19±\sqrt{361-240}}{2\times 10}$$

Add 361 to -240.

$$x=\frac{-19±\sqrt{121}}{2\times 10}$$

Take the square root of 121.

$$x=\frac{-19±11}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-19±11}{20}$$

Now solve the equation x=\frac{-19±11}{20} when ± is plus. Add -19 to 11.

$$x=-\frac{8}{20}$$

Reduce the fraction \frac{-8}{20} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-19±11}{20} when ± is minus. Subtract 11 from -19.

$$x=-\frac{30}{20}$$

Reduce the fraction \frac{-30}{20} to lowest terms by extracting and canceling out 10.

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

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

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

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

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

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

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

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

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

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

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

Multiply 5 times 2.

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

Cancel out 10, the greatest common factor in 10 and 10.

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

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 10

$$x ^ 2 +\frac{19}{10}x +\frac{3}{5} = 0$$

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

$$r + s = -\frac{19}{10} rs = \frac{3}{5}$$

Two numbers r and s sum up to -\frac{19}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{19}{10} = -\frac{19}{20}. 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>

$$r = -\frac{19}{20} - u s = -\frac{19}{20} + u$$

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

$$(-\frac{19}{20} - u) (-\frac{19}{20} + u) = \frac{3}{5}$$

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

$$\frac{361}{400} - u^2 = \frac{3}{5}$$

Simplify the expression by subtracting \frac{361}{400} on both sides

$$-u^2 = \frac{3}{5}-\frac{361}{400} = -\frac{121}{400}$$

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

$$u^2 = \frac{121}{400} u = \pm\sqrt{\frac{121}{400}} = \pm \frac{11}{20}$$

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

$$r =-\frac{19}{20} - \frac{11}{20} = -1.500 s = -\frac{19}{20} + \frac{11}{20} = -0.400$$

Evaluate

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