Solve 8z^2+14z+3=

$$8 z ^ { 2 } + 14 z + 3 =$$

Factor

$$\left(2z+3\right)\left(4z+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 8z^{2}+az+bz+3. To find a and b, set up a system to be solved.

$$a+b=14 ab=8\times 3=24$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 14.

$$a=2 b=12$$

Rewrite 8z^{2}+14z+3 as \left(8z^{2}+2z\right)+\left(12z+3\right).

$$\left(8z^{2}+2z\right)+\left(12z+3\right)$$

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

$$2z\left(4z+1\right)+3\left(4z+1\right)$$

Factor out common term 4z+1 by using distributive property.

$$\left(4z+1\right)\left(2z+3\right)$$

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.

$$8z^{2}+14z+3=0$$

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.

$$z=\frac{-14±\sqrt{14^{2}-4\times 8\times 3}}{2\times 8}$$

Square 14.

$$z=\frac{-14±\sqrt{196-4\times 8\times 3}}{2\times 8}$$

Multiply -4 times 8.

$$z=\frac{-14±\sqrt{196-32\times 3}}{2\times 8}$$

Multiply -32 times 3.

$$z=\frac{-14±\sqrt{196-96}}{2\times 8}$$

Add 196 to -96.

$$z=\frac{-14±\sqrt{100}}{2\times 8}$$

Take the square root of 100.

$$z=\frac{-14±10}{2\times 8}$$

Multiply 2 times 8.

$$z=\frac{-14±10}{16}$$

Now solve the equation z=\frac{-14±10}{16} when ± is plus. Add -14 to 10.

$$z=-\frac{4}{16}$$

Reduce the fraction \frac{-4}{16} to lowest terms by extracting and canceling out 4.

$$z=-\frac{1}{4}$$

Now solve the equation z=\frac{-14±10}{16} when ± is minus. Subtract 10 from -14.

$$z=-\frac{24}{16}$$

Reduce the fraction \frac{-24}{16} to lowest terms by extracting and canceling out 8.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{4} for x_{1} and -\frac{3}{2} for x_{2}.

$$8z^{2}+14z+3=8\left(z-\left(-\frac{1}{4}\right)\right)\left(z-\left(-\frac{3}{2}\right)\right)$$

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

$$8z^{2}+14z+3=8\left(z+\frac{1}{4}\right)\left(z+\frac{3}{2}\right)$$

Add \frac{1}{4} to z by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$8z^{2}+14z+3=8\times \frac{4z+1}{4}\left(z+\frac{3}{2}\right)$$

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

$$8z^{2}+14z+3=8\times \frac{4z+1}{4}\times \frac{2z+3}{2}$$

Multiply \frac{4z+1}{4} times \frac{2z+3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$8z^{2}+14z+3=8\times \frac{\left(4z+1\right)\left(2z+3\right)}{4\times 2}$$

Multiply 4 times 2.

$$8z^{2}+14z+3=8\times \frac{\left(4z+1\right)\left(2z+3\right)}{8}$$

Cancel out 8, the greatest common factor in 8 and 8.

$$8z^{2}+14z+3=\left(4z+1\right)\left(2z+3\right)$$

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 8

$$x ^ 2 +\frac{7}{4}x +\frac{3}{8} = 0$$

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

$$r + s = -\frac{7}{4} rs = \frac{3}{8}$$

Two numbers r and s sum up to -\frac{7}{4} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{4} = -\frac{7}{8}. 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>

$$r = -\frac{7}{8} - u s = -\frac{7}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{8}

$$(-\frac{7}{8} - u) (-\frac{7}{8} + u) = \frac{3}{8}$$

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

$$\frac{49}{64} - u^2 = \frac{3}{8}$$

Simplify the expression by subtracting \frac{49}{64} on both sides

$$-u^2 = \frac{3}{8}-\frac{49}{64} = -\frac{25}{64}$$

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

$$u^2 = \frac{25}{64} u = \pm\sqrt{\frac{25}{64}} = \pm \frac{5}{8}$$

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

$$r =-\frac{7}{8} - \frac{5}{8} = -1.500 s = -\frac{7}{8} + \frac{5}{8} = -0.250$$

Evaluate

$$\left(2z+3\right)\left(4z+1\right)$$