Solve 10a^2-13a-3

$$10 a ^ { 2 } - 13 a - 3$$

Factor

$$\left(2a-3\right)\left(5a+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10a^{2}+pa+qa-3. To find p and q, set up a system to be solved.

$$p+q=-13 pq=10\left(-3\right)=-30$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -13.

$$p=-15 q=2$$

Rewrite 10a^{2}-13a-3 as \left(10a^{2}-15a\right)+\left(2a-3\right).

$$\left(10a^{2}-15a\right)+\left(2a-3\right)$$

Factor out 5a in 10a^{2}-15a.

$$5a\left(2a-3\right)+2a-3$$

Factor out common term 2a-3 by using distributive property.

$$\left(2a-3\right)\left(5a+1\right)$$

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.

$$10a^{2}-13a-3=0$$

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.

$$a=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 10\left(-3\right)}}{2\times 10}$$

Square -13.

$$a=\frac{-\left(-13\right)±\sqrt{169-4\times 10\left(-3\right)}}{2\times 10}$$

Multiply -4 times 10.

$$a=\frac{-\left(-13\right)±\sqrt{169-40\left(-3\right)}}{2\times 10}$$

Multiply -40 times -3.

$$a=\frac{-\left(-13\right)±\sqrt{169+120}}{2\times 10}$$

Add 169 to 120.

$$a=\frac{-\left(-13\right)±\sqrt{289}}{2\times 10}$$

Take the square root of 289.

$$a=\frac{-\left(-13\right)±17}{2\times 10}$$

The opposite of -13 is 13.

$$a=\frac{13±17}{2\times 10}$$

Multiply 2 times 10.

$$a=\frac{13±17}{20}$$

Now solve the equation a=\frac{13±17}{20} when ± is plus. Add 13 to 17.

$$a=\frac{30}{20}$$

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

$$a=\frac{3}{2}$$

Now solve the equation a=\frac{13±17}{20} when ± is minus. Subtract 17 from 13.

$$a=-\frac{4}{20}$$

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

$$a=-\frac{1}{5}$$

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

$$10a^{2}-13a-3=10\left(a-\frac{3}{2}\right)\left(a-\left(-\frac{1}{5}\right)\right)$$

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

$$10a^{2}-13a-3=10\left(a-\frac{3}{2}\right)\left(a+\frac{1}{5}\right)$$

Subtract \frac{3}{2} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10a^{2}-13a-3=10\times \frac{2a-3}{2}\left(a+\frac{1}{5}\right)$$

Add \frac{1}{5} to a by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$10a^{2}-13a-3=10\times \frac{2a-3}{2}\times \frac{5a+1}{5}$$

Multiply \frac{2a-3}{2} times \frac{5a+1}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$10a^{2}-13a-3=10\times \frac{\left(2a-3\right)\left(5a+1\right)}{2\times 5}$$

Multiply 2 times 5.

$$10a^{2}-13a-3=10\times \frac{\left(2a-3\right)\left(5a+1\right)}{10}$$

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

$$10a^{2}-13a-3=\left(2a-3\right)\left(5a+1\right)$$

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 10

$$x ^ 2 -\frac{13}{10}x -\frac{3}{10} = 0$$

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

$$r + s = \frac{13}{10} rs = -\frac{3}{10}$$

Two numbers r and s sum up to \frac{13}{10} exactly when the average of the two numbers is \frac{1}{2}*\frac{13}{10} = \frac{13}{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>

$$r = \frac{13}{20} - u s = \frac{13}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{10}

$$(\frac{13}{20} - u) (\frac{13}{20} + u) = -\frac{3}{10}$$

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

$$\frac{169}{400} - u^2 = -\frac{3}{10}$$

Simplify the expression by subtracting \frac{169}{400} on both sides

$$-u^2 = -\frac{3}{10}-\frac{169}{400} = -\frac{289}{400}$$

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

$$u^2 = \frac{289}{400} u = \pm\sqrt{\frac{289}{400}} = \pm \frac{17}{20}$$

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

$$r =\frac{13}{20} - \frac{17}{20} = -0.200 s = \frac{13}{20} + \frac{17}{20} = 1.500$$

Evaluate

$$\left(2a-3\right)\left(5a+1\right)$$