Solve 14z^2-53z+14

$$14 z ^ { 2 } - 53 z + 14$$

Factor

$$\left(2z-7\right)\left(7z-2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 14z^{2}+az+bz+14. To find a and b, set up a system to be solved.

$$a+b=-53 ab=14\times 14=196$$

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 196.

$$-1,-196 -2,-98 -4,-49 -7,-28 -14,-14$$

Calculate the sum for each pair.

$$-1-196=-197 -2-98=-100 -4-49=-53 -7-28=-35 -14-14=-28$$

The solution is the pair that gives sum -53.

$$a=-49 b=-4$$

Rewrite 14z^{2}-53z+14 as \left(14z^{2}-49z\right)+\left(-4z+14\right).

$$\left(14z^{2}-49z\right)+\left(-4z+14\right)$$

Factor out 7z in the first and -2 in the second group.

$$7z\left(2z-7\right)-2\left(2z-7\right)$$

Factor out common term 2z-7 by using distributive property.

$$\left(2z-7\right)\left(7z-2\right)$$

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.

$$14z^{2}-53z+14=0$$

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.

$$z=\frac{-\left(-53\right)±\sqrt{\left(-53\right)^{2}-4\times 14\times 14}}{2\times 14}$$

Square -53.

$$z=\frac{-\left(-53\right)±\sqrt{2809-4\times 14\times 14}}{2\times 14}$$

Multiply -4 times 14.

$$z=\frac{-\left(-53\right)±\sqrt{2809-56\times 14}}{2\times 14}$$

Multiply -56 times 14.

$$z=\frac{-\left(-53\right)±\sqrt{2809-784}}{2\times 14}$$

Add 2809 to -784.

$$z=\frac{-\left(-53\right)±\sqrt{2025}}{2\times 14}$$

Take the square root of 2025.

$$z=\frac{-\left(-53\right)±45}{2\times 14}$$

The opposite of -53 is 53.

$$z=\frac{53±45}{2\times 14}$$

Multiply 2 times 14.

$$z=\frac{53±45}{28}$$

Now solve the equation z=\frac{53±45}{28} when ± is plus. Add 53 to 45.

$$z=\frac{98}{28}$$

Reduce the fraction \frac{98}{28} to lowest terms by extracting and canceling out 14.

$$z=\frac{7}{2}$$

Now solve the equation z=\frac{53±45}{28} when ± is minus. Subtract 45 from 53.

$$z=\frac{8}{28}$$

Reduce the fraction \frac{8}{28} to lowest terms by extracting and canceling out 4.

$$z=\frac{2}{7}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7}{2} for x_{1} and \frac{2}{7} for x_{2}.

$$14z^{2}-53z+14=14\left(z-\frac{7}{2}\right)\left(z-\frac{2}{7}\right)$$

Subtract \frac{7}{2} from z by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$14z^{2}-53z+14=14\times \frac{2z-7}{2}\left(z-\frac{2}{7}\right)$$

Subtract \frac{2}{7} from z by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$14z^{2}-53z+14=14\times \frac{2z-7}{2}\times \frac{7z-2}{7}$$

Multiply \frac{2z-7}{2} times \frac{7z-2}{7} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$14z^{2}-53z+14=14\times \frac{\left(2z-7\right)\left(7z-2\right)}{2\times 7}$$

Multiply 2 times 7.

$$14z^{2}-53z+14=14\times \frac{\left(2z-7\right)\left(7z-2\right)}{14}$$

Cancel out 14, the greatest common factor in 14 and 14.

$$14z^{2}-53z+14=\left(2z-7\right)\left(7z-2\right)$$

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 14

$$x ^ 2 -\frac{53}{14}x +1 = 0$$

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

$$r + s = \frac{53}{14} rs = 1$$

Two numbers r and s sum up to \frac{53}{14} exactly when the average of the two numbers is \frac{1}{2}*\frac{53}{14} = \frac{53}{28}. 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>

$$r = \frac{53}{28} - u s = \frac{53}{28} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 1

$$(\frac{53}{28} - u) (\frac{53}{28} + u) = 1$$

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

$$\frac{2809}{784} - u^2 = 1$$

Simplify the expression by subtracting \frac{2809}{784} on both sides

$$-u^2 = 1-\frac{2809}{784} = -\frac{2025}{784}$$

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

$$u^2 = \frac{2025}{784} u = \pm\sqrt{\frac{2025}{784}} = \pm \frac{45}{28}$$

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

$$r =\frac{53}{28} - \frac{45}{28} = 0.286 s = \frac{53}{28} + \frac{45}{28} = 3.500$$

Evaluate

$$\left(2z-7\right)\left(7z-2\right)$$