Solve 15x^2+25x+10

$$15 x ^ { 2 } + 25 x + 10$$

Factor

$$5\left(x+1\right)\left(3x+2\right)$$

Solution Steps

Factor out 5.

$$5\left(3x^{2}+5x+2\right)$$

Consider 3x^{2}+5x+2. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=5 ab=3\times 2=6$$

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 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 5.

$$a=2 b=3$$

Rewrite 3x^{2}+5x+2 as \left(3x^{2}+2x\right)+\left(3x+2\right).

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

Factor out x in 3x^{2}+2x.

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

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

$$\left(3x+2\right)\left(x+1\right)$$

Rewrite the complete factored expression.

$$5\left(3x+2\right)\left(x+1\right)$$

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.

$$15x^{2}+25x+10=0$$

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.

$$x=\frac{-25±\sqrt{25^{2}-4\times 15\times 10}}{2\times 15}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\times 15\times 10}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-25±\sqrt{625-60\times 10}}{2\times 15}$$

Multiply -60 times 10.

$$x=\frac{-25±\sqrt{625-600}}{2\times 15}$$

Add 625 to -600.

$$x=\frac{-25±\sqrt{25}}{2\times 15}$$

Take the square root of 25.

$$x=\frac{-25±5}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{-25±5}{30}$$

Now solve the equation x=\frac{-25±5}{30} when ± is plus. Add -25 to 5.

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

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

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

Now solve the equation x=\frac{-25±5}{30} when ± is minus. Subtract 5 from -25.

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

Divide -30 by 30.

$$x=-1$$

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 -1 for x_{2}.

$$15x^{2}+25x+10=15\left(x-\left(-\frac{2}{3}\right)\right)\left(x-\left(-1\right)\right)$$

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

$$15x^{2}+25x+10=15\left(x+\frac{2}{3}\right)\left(x+1\right)$$

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

$$15x^{2}+25x+10=15\times \frac{3x+2}{3}\left(x+1\right)$$

Cancel out 3, the greatest common factor in 15 and 3.

$$15x^{2}+25x+10=5\left(3x+2\right)\left(x+1\right)$$

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 15

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

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

$$r + s = -\frac{5}{3} rs = \frac{2}{3}$$

Two numbers r and s sum up to -\frac{5}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{3} = -\frac{5}{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>

$$r = -\frac{5}{6} - u s = -\frac{5}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{3}

$$(-\frac{5}{6} - u) (-\frac{5}{6} + u) = \frac{2}{3}$$

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

$$\frac{25}{36} - u^2 = \frac{2}{3}$$

Simplify the expression by subtracting \frac{25}{36} on both sides

$$-u^2 = \frac{2}{3}-\frac{25}{36} = -\frac{1}{36}$$

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

$$u^2 = \frac{1}{36} u = \pm\sqrt{\frac{1}{36}} = \pm \frac{1}{6}$$

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

$$r =-\frac{5}{6} - \frac{1}{6} = -1.000 s = -\frac{5}{6} + \frac{1}{6} = -0.667$$

Evaluate

$$5\left(x+1\right)\left(3x+2\right)$$