Solve 2a^2-3a-5

$$2 a ^ { 2 } - 3 a - 5$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2a^{2}+pa+qa-5. To find p and q, set up a system to be solved.

$$p+q=-3 pq=2\left(-5\right)=-10$$

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 -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$p=-5 q=2$$

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

$$\left(2a^{2}-5a\right)+\left(2a-5\right)$$

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

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

Factor out common term 2a-5 by using distributive property.

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

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.

$$2a^{2}-3a-5=0$$

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.

$$a=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\left(-5\right)}}{2\times 2}$$

Square -3.

$$a=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-5\right)}}{2\times 2}$$

Multiply -4 times 2.

$$a=\frac{-\left(-3\right)±\sqrt{9-8\left(-5\right)}}{2\times 2}$$

Multiply -8 times -5.

$$a=\frac{-\left(-3\right)±\sqrt{9+40}}{2\times 2}$$

Add 9 to 40.

$$a=\frac{-\left(-3\right)±\sqrt{49}}{2\times 2}$$

Take the square root of 49.

$$a=\frac{-\left(-3\right)±7}{2\times 2}$$

The opposite of -3 is 3.

$$a=\frac{3±7}{2\times 2}$$

Multiply 2 times 2.

$$a=\frac{3±7}{4}$$

Now solve the equation a=\frac{3±7}{4} when ± is plus. Add 3 to 7.

$$a=\frac{10}{4}$$

Reduce the fraction \frac{10}{4} to lowest terms by extracting and canceling out 2.

$$a=\frac{5}{2}$$

Now solve the equation a=\frac{3±7}{4} when ± is minus. Subtract 7 from 3.

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

Divide -4 by 4.

$$a=-1$$

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

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

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

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

Subtract \frac{5}{2} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Cancel out 2, the greatest common factor in 2 and 2.

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

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 2

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

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

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

Two numbers r and s sum up to \frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{2} = \frac{3}{4}. 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>

$$r = \frac{3}{4} - u s = \frac{3}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{2}

$$(\frac{3}{4} - u) (\frac{3}{4} + u) = -\frac{5}{2}$$

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

$$\frac{9}{16} - u^2 = -\frac{5}{2}$$

Simplify the expression by subtracting \frac{9}{16} on both sides

$$-u^2 = -\frac{5}{2}-\frac{9}{16} = -\frac{49}{16}$$

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

$$u^2 = \frac{49}{16} u = \pm\sqrt{\frac{49}{16}} = \pm \frac{7}{4}$$

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

$$r =\frac{3}{4} - \frac{7}{4} = -1 s = \frac{3}{4} + \frac{7}{4} = 2.500$$

Evaluate

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