Solve 12z^2+32z-35

$$12 z ^ { 2 } + 32 z - 35$$

Factor

$$\left(6z-5\right)\left(2z+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12z^{2}+az+bz-35. To find a and b, set up a system to be solved.

$$a+b=32 ab=12\left(-35\right)=-420$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -420.

$$-1,420 -2,210 -3,140 -4,105 -5,84 -6,70 -7,60 -10,42 -12,35 -14,30 -15,28 -20,21$$

Calculate the sum for each pair.

$$-1+420=419 -2+210=208 -3+140=137 -4+105=101 -5+84=79 -6+70=64 -7+60=53 -10+42=32 -12+35=23 -14+30=16 -15+28=13 -20+21=1$$

The solution is the pair that gives sum 32.

$$a=-10 b=42$$

Rewrite 12z^{2}+32z-35 as \left(12z^{2}-10z\right)+\left(42z-35\right).

$$\left(12z^{2}-10z\right)+\left(42z-35\right)$$

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

$$2z\left(6z-5\right)+7\left(6z-5\right)$$

Factor out common term 6z-5 by using distributive property.

$$\left(6z-5\right)\left(2z+7\right)$$

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.

$$12z^{2}+32z-35=0$$

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.

$$z=\frac{-32±\sqrt{32^{2}-4\times 12\left(-35\right)}}{2\times 12}$$

Square 32.

$$z=\frac{-32±\sqrt{1024-4\times 12\left(-35\right)}}{2\times 12}$$

Multiply -4 times 12.

$$z=\frac{-32±\sqrt{1024-48\left(-35\right)}}{2\times 12}$$

Multiply -48 times -35.

$$z=\frac{-32±\sqrt{1024+1680}}{2\times 12}$$

Add 1024 to 1680.

$$z=\frac{-32±\sqrt{2704}}{2\times 12}$$

Take the square root of 2704.

$$z=\frac{-32±52}{2\times 12}$$

Multiply 2 times 12.

$$z=\frac{-32±52}{24}$$

Now solve the equation z=\frac{-32±52}{24} when ± is plus. Add -32 to 52.

$$z=\frac{20}{24}$$

Reduce the fraction \frac{20}{24} to lowest terms by extracting and canceling out 4.

$$z=\frac{5}{6}$$

Now solve the equation z=\frac{-32±52}{24} when ± is minus. Subtract 52 from -32.

$$z=-\frac{84}{24}$$

Reduce the fraction \frac{-84}{24} to lowest terms by extracting and canceling out 12.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{6} for x_{1} and -\frac{7}{2} for x_{2}.

$$12z^{2}+32z-35=12\left(z-\frac{5}{6}\right)\left(z-\left(-\frac{7}{2}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$12z^{2}+32z-35=12\left(z-\frac{5}{6}\right)\left(z+\frac{7}{2}\right)$$

Subtract \frac{5}{6} from z by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12z^{2}+32z-35=12\times \frac{6z-5}{6}\left(z+\frac{7}{2}\right)$$

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

$$12z^{2}+32z-35=12\times \frac{6z-5}{6}\times \frac{2z+7}{2}$$

Multiply \frac{6z-5}{6} times \frac{2z+7}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12z^{2}+32z-35=12\times \frac{\left(6z-5\right)\left(2z+7\right)}{6\times 2}$$

Multiply 6 times 2.

$$12z^{2}+32z-35=12\times \frac{\left(6z-5\right)\left(2z+7\right)}{12}$$

Cancel out 12, the greatest common factor in 12 and 12.

$$12z^{2}+32z-35=\left(6z-5\right)\left(2z+7\right)$$

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 12

$$x ^ 2 +\frac{8}{3}x -\frac{35}{12} = 0$$

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

$$r + s = -\frac{8}{3} rs = -\frac{35}{12}$$

Two numbers r and s sum up to -\frac{8}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{8}{3} = -\frac{4}{3}. 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>

$$r = -\frac{4}{3} - u s = -\frac{4}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{35}{12}

$$(-\frac{4}{3} - u) (-\frac{4}{3} + u) = -\frac{35}{12}$$

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

$$\frac{16}{9} - u^2 = -\frac{35}{12}$$

Simplify the expression by subtracting \frac{16}{9} on both sides

$$-u^2 = -\frac{35}{12}-\frac{16}{9} = -\frac{169}{36}$$

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

$$u^2 = \frac{169}{36} u = \pm\sqrt{\frac{169}{36}} = \pm \frac{13}{6}$$

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

$$r =-\frac{4}{3} - \frac{13}{6} = -3.500 s = -\frac{4}{3} + \frac{13}{6} = 0.833$$

Evaluate

$$\left(6z-5\right)\left(2z+7\right)$$