Solve 2z^2-23z+30

$$2 z ^ { 2 } - 23 z + 30$$

Factor

$$\left(z-10\right)\left(2z-3\right)$$

Solution Steps

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

$$a+b=-23 ab=2\times 30=60$$

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 60.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Calculate the sum for each pair.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

The solution is the pair that gives sum -23.

$$a=-20 b=-3$$

Rewrite 2z^{2}-23z+30 as \left(2z^{2}-20z\right)+\left(-3z+30\right).

$$\left(2z^{2}-20z\right)+\left(-3z+30\right)$$

Factor out 2z in the first and -3 in the second group.

$$2z\left(z-10\right)-3\left(z-10\right)$$

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

$$\left(z-10\right)\left(2z-3\right)$$

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.

$$2z^{2}-23z+30=0$$

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.

$$z=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 2\times 30}}{2\times 2}$$

Square -23.

$$z=\frac{-\left(-23\right)±\sqrt{529-4\times 2\times 30}}{2\times 2}$$

Multiply -4 times 2.

$$z=\frac{-\left(-23\right)±\sqrt{529-8\times 30}}{2\times 2}$$

Multiply -8 times 30.

$$z=\frac{-\left(-23\right)±\sqrt{529-240}}{2\times 2}$$

Add 529 to -240.

$$z=\frac{-\left(-23\right)±\sqrt{289}}{2\times 2}$$

Take the square root of 289.

$$z=\frac{-\left(-23\right)±17}{2\times 2}$$

The opposite of -23 is 23.

$$z=\frac{23±17}{2\times 2}$$

Multiply 2 times 2.

$$z=\frac{23±17}{4}$$

Now solve the equation z=\frac{23±17}{4} when ± is plus. Add 23 to 17.

$$z=\frac{40}{4}$$

Divide 40 by 4.

$$z=10$$

Now solve the equation z=\frac{23±17}{4} when ± is minus. Subtract 17 from 23.

$$z=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

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

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 \frac{3}{2} for x_{2}.

$$2z^{2}-23z+30=2\left(z-10\right)\left(z-\frac{3}{2}\right)$$

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

$$2z^{2}-23z+30=2\left(z-10\right)\times \frac{2z-3}{2}$$

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

$$2z^{2}-23z+30=\left(z-10\right)\left(2z-3\right)$$

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.This is achieved by dividing both sides of the equation by 2

$$x ^ 2 -\frac{23}{2}x +15 = 0$$

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

$$r + s = \frac{23}{2} rs = 15$$

Two numbers r and s sum up to \frac{23}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{23}{2} = \frac{23}{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>

$$r = \frac{23}{4} - u s = \frac{23}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 15

$$(\frac{23}{4} - u) (\frac{23}{4} + u) = 15$$

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

$$\frac{529}{16} - u^2 = 15$$

Simplify the expression by subtracting \frac{529}{16} on both sides

$$-u^2 = 15-\frac{529}{16} = -\frac{289}{16}$$

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

$$u^2 = \frac{289}{16} u = \pm\sqrt{\frac{289}{16}} = \pm \frac{17}{4}$$

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

$$r =\frac{23}{4} - \frac{17}{4} = 1.500 s = \frac{23}{4} + \frac{17}{4} = 10$$

Evaluate

$$\left(z-10\right)\left(2z-3\right)$$