Solve 13x^2+21x-10

$$13 x ^ { 2 } + 21 x - 10$$

Factor

$$\left(13x-5\right)\left(x+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 13x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=21 ab=13\left(-10\right)=-130$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -130.

$$-1,130 -2,65 -5,26 -10,13$$

Calculate the sum for each pair.

$$-1+130=129 -2+65=63 -5+26=21 -10+13=3$$

The solution is the pair that gives sum 21.

$$a=-5 b=26$$

Rewrite 13x^{2}+21x-10 as \left(13x^{2}-5x\right)+\left(26x-10\right).

$$\left(13x^{2}-5x\right)+\left(26x-10\right)$$

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

$$x\left(13x-5\right)+2\left(13x-5\right)$$

Factor out common term 13x-5 by using distributive property.

$$\left(13x-5\right)\left(x+2\right)$$

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.

$$13x^{2}+21x-10=0$$

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.

$$x=\frac{-21±\sqrt{21^{2}-4\times 13\left(-10\right)}}{2\times 13}$$

Square 21.

$$x=\frac{-21±\sqrt{441-4\times 13\left(-10\right)}}{2\times 13}$$

Multiply -4 times 13.

$$x=\frac{-21±\sqrt{441-52\left(-10\right)}}{2\times 13}$$

Multiply -52 times -10.

$$x=\frac{-21±\sqrt{441+520}}{2\times 13}$$

Add 441 to 520.

$$x=\frac{-21±\sqrt{961}}{2\times 13}$$

Take the square root of 961.

$$x=\frac{-21±31}{2\times 13}$$

Multiply 2 times 13.

$$x=\frac{-21±31}{26}$$

Now solve the equation x=\frac{-21±31}{26} when ± is plus. Add -21 to 31.

$$x=\frac{10}{26}$$

Reduce the fraction \frac{10}{26} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{13}$$

Now solve the equation x=\frac{-21±31}{26} when ± is minus. Subtract 31 from -21.

$$x=-\frac{52}{26}$$

Divide -52 by 26.

$$x=-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{13} for x_{1} and -2 for x_{2}.

$$13x^{2}+21x-10=13\left(x-\frac{5}{13}\right)\left(x-\left(-2\right)\right)$$

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

$$13x^{2}+21x-10=13\left(x-\frac{5}{13}\right)\left(x+2\right)$$

Subtract \frac{5}{13} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$13x^{2}+21x-10=13\times \frac{13x-5}{13}\left(x+2\right)$$

Cancel out 13, the greatest common factor in 13 and 13.

$$13x^{2}+21x-10=\left(13x-5\right)\left(x+2\right)$$

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 13

$$x ^ 2 +\frac{21}{13}x -\frac{10}{13} = 0$$

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

$$r + s = -\frac{21}{13} rs = -\frac{10}{13}$$

Two numbers r and s sum up to -\frac{21}{13} exactly when the average of the two numbers is \frac{1}{2}*-\frac{21}{13} = -\frac{21}{26}. 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>

$$r = -\frac{21}{26} - u s = -\frac{21}{26} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{10}{13}

$$(-\frac{21}{26} - u) (-\frac{21}{26} + u) = -\frac{10}{13}$$

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

$$\frac{441}{676} - u^2 = -\frac{10}{13}$$

Simplify the expression by subtracting \frac{441}{676} on both sides

$$-u^2 = -\frac{10}{13}-\frac{441}{676} = -\frac{961}{676}$$

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

$$u^2 = \frac{961}{676} u = \pm\sqrt{\frac{961}{676}} = \pm \frac{31}{26}$$

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

$$r =-\frac{21}{26} - \frac{31}{26} = -2 s = -\frac{21}{26} + \frac{31}{26} = 0.385$$

Evaluate

$$\left(13x-5\right)\left(x+2\right)$$