Solve 6z^2-13z+5

$$6 z ^ { 2 } - 13 z + 5$$

Factor

$$\left(2z-1\right)\left(3z-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6z^{2}+az+bz+5. To find a and b, set up a system to be solved.

$$a+b=-13 ab=6\times 5=30$$

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 6z^{2}-13z+5 as \left(6z^{2}-10z\right)+\left(-3z+5\right).

$$\left(6z^{2}-10z\right)+\left(-3z+5\right)$$

Factor out 2z in the first and -1 in the second group.

$$2z\left(3z-5\right)-\left(3z-5\right)$$

Factor out common term 3z-5 by using distributive property.

$$\left(3z-5\right)\left(2z-1\right)$$

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.

$$6z^{2}-13z+5=0$$

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.

$$z=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 6\times 5}}{2\times 6}$$

Square -13.

$$z=\frac{-\left(-13\right)±\sqrt{169-4\times 6\times 5}}{2\times 6}$$

Multiply -4 times 6.

$$z=\frac{-\left(-13\right)±\sqrt{169-24\times 5}}{2\times 6}$$

Multiply -24 times 5.

$$z=\frac{-\left(-13\right)±\sqrt{169-120}}{2\times 6}$$

Add 169 to -120.

$$z=\frac{-\left(-13\right)±\sqrt{49}}{2\times 6}$$

Take the square root of 49.

$$z=\frac{-\left(-13\right)±7}{2\times 6}$$

The opposite of -13 is 13.

$$z=\frac{13±7}{2\times 6}$$

Multiply 2 times 6.

$$z=\frac{13±7}{12}$$

Now solve the equation z=\frac{13±7}{12} when ± is plus. Add 13 to 7.

$$z=\frac{20}{12}$$

Reduce the fraction \frac{20}{12} to lowest terms by extracting and canceling out 4.

$$z=\frac{5}{3}$$

Now solve the equation z=\frac{13±7}{12} when ± is minus. Subtract 7 from 13.

$$z=\frac{6}{12}$$

Reduce the fraction \frac{6}{12} to lowest terms by extracting and canceling out 6.

$$z=\frac{1}{2}$$

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

$$6z^{2}-13z+5=6\left(z-\frac{5}{3}\right)\left(z-\frac{1}{2}\right)$$

Subtract \frac{5}{3} from z by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6z^{2}-13z+5=6\times \frac{3z-5}{3}\left(z-\frac{1}{2}\right)$$

Subtract \frac{1}{2} from z by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6z^{2}-13z+5=6\times \frac{3z-5}{3}\times \frac{2z-1}{2}$$

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

$$6z^{2}-13z+5=6\times \frac{\left(3z-5\right)\left(2z-1\right)}{3\times 2}$$

Multiply 3 times 2.

$$6z^{2}-13z+5=6\times \frac{\left(3z-5\right)\left(2z-1\right)}{6}$$

Cancel out 6, the greatest common factor in 6 and 6.

$$6z^{2}-13z+5=\left(3z-5\right)\left(2z-1\right)$$

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 6

$$x ^ 2 -\frac{13}{6}x +\frac{5}{6} = 0$$

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

$$r + s = \frac{13}{6} rs = \frac{5}{6}$$

Two numbers r and s sum up to \frac{13}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{13}{6} = \frac{13}{12}. 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>

$$r = \frac{13}{12} - u s = \frac{13}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{6}

$$(\frac{13}{12} - u) (\frac{13}{12} + u) = \frac{5}{6}$$

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

$$\frac{169}{144} - u^2 = \frac{5}{6}$$

Simplify the expression by subtracting \frac{169}{144} on both sides

$$-u^2 = \frac{5}{6}-\frac{169}{144} = -\frac{49}{144}$$

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

$$u^2 = \frac{49}{144} u = \pm\sqrt{\frac{49}{144}} = \pm \frac{7}{12}$$

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

$$r =\frac{13}{12} - \frac{7}{12} = 0.500 s = \frac{13}{12} + \frac{7}{12} = 1.667$$

Evaluate

$$\left(2z-1\right)\left(3z-5\right)$$