Solve 8y^2+11y+3

$$8 y ^ { 2 } + 11 y + 3$$

Factor

$$\left(y+1\right)\left(8y+3\right)$$

Solution Steps

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

$$a+b=11 ab=8\times 3=24$$

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 11.

$$a=3 b=8$$

Rewrite 8y^{2}+11y+3 as \left(8y^{2}+3y\right)+\left(8y+3\right).

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

Factor out y in 8y^{2}+3y.

$$y\left(8y+3\right)+8y+3$$

Factor out common term 8y+3 by using distributive property.

$$\left(8y+3\right)\left(y+1\right)$$

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.

$$8y^{2}+11y+3=0$$

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.

$$y=\frac{-11±\sqrt{11^{2}-4\times 8\times 3}}{2\times 8}$$

Square 11.

$$y=\frac{-11±\sqrt{121-4\times 8\times 3}}{2\times 8}$$

Multiply -4 times 8.

$$y=\frac{-11±\sqrt{121-32\times 3}}{2\times 8}$$

Multiply -32 times 3.

$$y=\frac{-11±\sqrt{121-96}}{2\times 8}$$

Add 121 to -96.

$$y=\frac{-11±\sqrt{25}}{2\times 8}$$

Take the square root of 25.

$$y=\frac{-11±5}{2\times 8}$$

Multiply 2 times 8.

$$y=\frac{-11±5}{16}$$

Now solve the equation y=\frac{-11±5}{16} when ± is plus. Add -11 to 5.

$$y=-\frac{6}{16}$$

Reduce the fraction \frac{-6}{16} to lowest terms by extracting and canceling out 2.

$$y=-\frac{3}{8}$$

Now solve the equation y=\frac{-11±5}{16} when ± is minus. Subtract 5 from -11.

$$y=-\frac{16}{16}$$

Divide -16 by 16.

$$y=-1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{3}{8} for x_{1} and -1 for x_{2}.

$$8y^{2}+11y+3=8\left(y-\left(-\frac{3}{8}\right)\right)\left(y-\left(-1\right)\right)$$

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

$$8y^{2}+11y+3=8\left(y+\frac{3}{8}\right)\left(y+1\right)$$

Add \frac{3}{8} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$8y^{2}+11y+3=8\times \frac{8y+3}{8}\left(y+1\right)$$

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

$$8y^{2}+11y+3=\left(8y+3\right)\left(y+1\right)$$

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 8

$$x ^ 2 +\frac{11}{8}x +\frac{3}{8} = 0$$

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

$$r + s = -\frac{11}{8} rs = \frac{3}{8}$$

Two numbers r and s sum up to -\frac{11}{8} exactly when the average of the two numbers is \frac{1}{2}*-\frac{11}{8} = -\frac{11}{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>

$$r = -\frac{11}{16} - u s = -\frac{11}{16} + u$$

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

$$(-\frac{11}{16} - u) (-\frac{11}{16} + u) = \frac{3}{8}$$

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

$$\frac{121}{256} - u^2 = \frac{3}{8}$$

Simplify the expression by subtracting \frac{121}{256} on both sides

$$-u^2 = \frac{3}{8}-\frac{121}{256} = -\frac{25}{256}$$

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

$$u^2 = \frac{25}{256} u = \pm\sqrt{\frac{25}{256}} = \pm \frac{5}{16}$$

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

$$r =-\frac{11}{16} - \frac{5}{16} = -1 s = -\frac{11}{16} + \frac{5}{16} = -0.375$$

Evaluate

$$\left(y+1\right)\left(8y+3\right)$$