Solve 16z^2-54z+35

$$16 z ^ { 2 } - 54 z + 35$$

Factor

$$\left(2z-5\right)\left(8z-7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 16z^{2}+az+bz+35. To find a and b, set up a system to be solved.

$$a+b=-54 ab=16\times 35=560$$

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 560.

$$-1,-560 -2,-280 -4,-140 -5,-112 -7,-80 -8,-70 -10,-56 -14,-40 -16,-35 -20,-28$$

Calculate the sum for each pair.

$$-1-560=-561 -2-280=-282 -4-140=-144 -5-112=-117 -7-80=-87 -8-70=-78 -10-56=-66 -14-40=-54 -16-35=-51 -20-28=-48$$

The solution is the pair that gives sum -54.

$$a=-40 b=-14$$

Rewrite 16z^{2}-54z+35 as \left(16z^{2}-40z\right)+\left(-14z+35\right).

$$\left(16z^{2}-40z\right)+\left(-14z+35\right)$$

Factor out 8z in the first and -7 in the second group.

$$8z\left(2z-5\right)-7\left(2z-5\right)$$

Factor out common term 2z-5 by using distributive property.

$$\left(2z-5\right)\left(8z-7\right)$$

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.

$$16z^{2}-54z+35=0$$

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.

$$z=\frac{-\left(-54\right)±\sqrt{\left(-54\right)^{2}-4\times 16\times 35}}{2\times 16}$$

Square -54.

$$z=\frac{-\left(-54\right)±\sqrt{2916-4\times 16\times 35}}{2\times 16}$$

Multiply -4 times 16.

$$z=\frac{-\left(-54\right)±\sqrt{2916-64\times 35}}{2\times 16}$$

Multiply -64 times 35.

$$z=\frac{-\left(-54\right)±\sqrt{2916-2240}}{2\times 16}$$

Add 2916 to -2240.

$$z=\frac{-\left(-54\right)±\sqrt{676}}{2\times 16}$$

Take the square root of 676.

$$z=\frac{-\left(-54\right)±26}{2\times 16}$$

The opposite of -54 is 54.

$$z=\frac{54±26}{2\times 16}$$

Multiply 2 times 16.

$$z=\frac{54±26}{32}$$

Now solve the equation z=\frac{54±26}{32} when ± is plus. Add 54 to 26.

$$z=\frac{80}{32}$$

Reduce the fraction \frac{80}{32} to lowest terms by extracting and canceling out 16.

$$z=\frac{5}{2}$$

Now solve the equation z=\frac{54±26}{32} when ± is minus. Subtract 26 from 54.

$$z=\frac{28}{32}$$

Reduce the fraction \frac{28}{32} to lowest terms by extracting and canceling out 4.

$$z=\frac{7}{8}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{2} for x_{1} and \frac{7}{8} for x_{2}.

$$16z^{2}-54z+35=16\left(z-\frac{5}{2}\right)\left(z-\frac{7}{8}\right)$$

Subtract \frac{5}{2} from z by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$16z^{2}-54z+35=16\times \frac{2z-5}{2}\left(z-\frac{7}{8}\right)$$

Subtract \frac{7}{8} from z by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$16z^{2}-54z+35=16\times \frac{2z-5}{2}\times \frac{8z-7}{8}$$

Multiply \frac{2z-5}{2} times \frac{8z-7}{8} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$16z^{2}-54z+35=16\times \frac{\left(2z-5\right)\left(8z-7\right)}{2\times 8}$$

Multiply 2 times 8.

$$16z^{2}-54z+35=16\times \frac{\left(2z-5\right)\left(8z-7\right)}{16}$$

Cancel out 16, the greatest common factor in 16 and 16.

$$16z^{2}-54z+35=\left(2z-5\right)\left(8z-7\right)$$

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 16

$$x ^ 2 -\frac{27}{8}x +\frac{35}{16} = 0$$

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

$$r + s = \frac{27}{8} rs = \frac{35}{16}$$

Two numbers r and s sum up to \frac{27}{8} exactly when the average of the two numbers is \frac{1}{2}*\frac{27}{8} = \frac{27}{16}. 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>

$$r = \frac{27}{16} - u s = \frac{27}{16} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{35}{16}

$$(\frac{27}{16} - u) (\frac{27}{16} + u) = \frac{35}{16}$$

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

$$\frac{729}{256} - u^2 = \frac{35}{16}$$

Simplify the expression by subtracting \frac{729}{256} on both sides

$$-u^2 = \frac{35}{16}-\frac{729}{256} = -\frac{169}{256}$$

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

$$u^2 = \frac{169}{256} u = \pm\sqrt{\frac{169}{256}} = \pm \frac{13}{16}$$

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

$$r =\frac{27}{16} - \frac{13}{16} = 0.875 s = \frac{27}{16} + \frac{13}{16} = 2.500$$

Evaluate

$$\left(2z-5\right)\left(8z-7\right)$$