Solve d^2-13d+30

$$d ^ { 2 } - 13 d + 30$$

Factor

$$\left(d-10\right)\left(d-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as d^{2}+ad+bd+30. To find a and b, set up a system to be solved.

$$a+b=-13 ab=1\times 30=30$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 30.

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

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -13.

$$a=-10 b=-3$$

Rewrite d^{2}-13d+30 as \left(d^{2}-10d\right)+\left(-3d+30\right).

$$\left(d^{2}-10d\right)+\left(-3d+30\right)$$

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

$$d\left(d-10\right)-3\left(d-10\right)$$

Factor out common term d-10 by using distributive property.

$$\left(d-10\right)\left(d-3\right)$$

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.

$$d^{2}-13d+30=0$$

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.

$$d=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 30}}{2}$$

Square -13.

$$d=\frac{-\left(-13\right)±\sqrt{169-4\times 30}}{2}$$

Multiply -4 times 30.

$$d=\frac{-\left(-13\right)±\sqrt{169-120}}{2}$$

Add 169 to -120.

$$d=\frac{-\left(-13\right)±\sqrt{49}}{2}$$

Take the square root of 49.

$$d=\frac{-\left(-13\right)±7}{2}$$

The opposite of -13 is 13.

$$d=\frac{13±7}{2}$$

Now solve the equation d=\frac{13±7}{2} when ± is plus. Add 13 to 7.

$$d=\frac{20}{2}$$

Divide 20 by 2.

$$d=10$$

Now solve the equation d=\frac{13±7}{2} when ± is minus. Subtract 7 from 13.

$$d=\frac{6}{2}$$

Divide 6 by 2.

$$d=3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 10 for x_{1} and 3 for x_{2}.

$$d^{2}-13d+30=\left(d-10\right)\left(d-3\right)$$

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.

$$x ^ 2 -13x +30 = 0$$

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

$$r + s = 13 rs = 30$$

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

$$r = \frac{13}{2} - u s = \frac{13}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 30

$$(\frac{13}{2} - u) (\frac{13}{2} + u) = 30$$

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

$$\frac{169}{4} - u^2 = 30$$

Simplify the expression by subtracting \frac{169}{4} on both sides

$$-u^2 = 30-\frac{169}{4} = -\frac{49}{4}$$

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

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

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

$$r =\frac{13}{2} - \frac{7}{2} = 3 s = \frac{13}{2} + \frac{7}{2} = 10$$

Evaluate

$$\left(d-10\right)\left(d-3\right)$$