Solve 3d^2-4d-15

$$3 d ^ { 2 } - 4 d - 15$$

Factor

$$\left(d-3\right)\left(3d+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3d^{2}+ad+bd-15. To find a and b, set up a system to be solved.

$$a+b=-4 ab=3\left(-15\right)=-45$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -45.

$$1,-45 3,-15 5,-9$$

Calculate the sum for each pair.

$$1-45=-44 3-15=-12 5-9=-4$$

The solution is the pair that gives sum -4.

$$a=-9 b=5$$

Rewrite 3d^{2}-4d-15 as \left(3d^{2}-9d\right)+\left(5d-15\right).

$$\left(3d^{2}-9d\right)+\left(5d-15\right)$$

Factor out 3d in the first and 5 in the second group.

$$3d\left(d-3\right)+5\left(d-3\right)$$

Factor out common term d-3 by using distributive property.

$$\left(d-3\right)\left(3d+5\right)$$

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.

$$3d^{2}-4d-15=0$$

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.

$$d=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 3\left(-15\right)}}{2\times 3}$$

Square -4.

$$d=\frac{-\left(-4\right)±\sqrt{16-4\times 3\left(-15\right)}}{2\times 3}$$

Multiply -4 times 3.

$$d=\frac{-\left(-4\right)±\sqrt{16-12\left(-15\right)}}{2\times 3}$$

Multiply -12 times -15.

$$d=\frac{-\left(-4\right)±\sqrt{16+180}}{2\times 3}$$

Add 16 to 180.

$$d=\frac{-\left(-4\right)±\sqrt{196}}{2\times 3}$$

Take the square root of 196.

$$d=\frac{-\left(-4\right)±14}{2\times 3}$$

The opposite of -4 is 4.

$$d=\frac{4±14}{2\times 3}$$

Multiply 2 times 3.

$$d=\frac{4±14}{6}$$

Now solve the equation d=\frac{4±14}{6} when ± is plus. Add 4 to 14.

$$d=\frac{18}{6}$$

Divide 18 by 6.

$$d=3$$

Now solve the equation d=\frac{4±14}{6} when ± is minus. Subtract 14 from 4.

$$d=-\frac{10}{6}$$

Reduce the fraction \frac{-10}{6} to lowest terms by extracting and canceling out 2.

$$d=-\frac{5}{3}$$

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

$$3d^{2}-4d-15=3\left(d-3\right)\left(d-\left(-\frac{5}{3}\right)\right)$$

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

$$3d^{2}-4d-15=3\left(d-3\right)\left(d+\frac{5}{3}\right)$$

Add \frac{5}{3} to d by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3d^{2}-4d-15=3\left(d-3\right)\times \frac{3d+5}{3}$$

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

$$3d^{2}-4d-15=\left(d-3\right)\left(3d+5\right)$$

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 3

$$x ^ 2 -\frac{4}{3}x -5 = 0$$

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

$$r + s = \frac{4}{3} rs = -5$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = -5

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

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

$$\frac{4}{9} - u^2 = -5$$

Simplify the expression by subtracting \frac{4}{9} on both sides

$$-u^2 = -5-\frac{4}{9} = -\frac{49}{9}$$

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

$$u^2 = \frac{49}{9} u = \pm\sqrt{\frac{49}{9}} = \pm \frac{7}{3}$$

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

$$r =\frac{2}{3} - \frac{7}{3} = -1.667 s = \frac{2}{3} + \frac{7}{3} = 3$$

Evaluate

$$\left(d-3\right)\left(3d+5\right)$$