Solve 4h^2+8h+3

$$4 h ^ { 2 } + 8 h + 3$$

Factor

$$\left(2h+1\right)\left(2h+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4h^{2}+ah+bh+3. To find a and b, set up a system to be solved.

$$a+b=8 ab=4\times 3=12$$

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 12.

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

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 8.

$$a=2 b=6$$

Rewrite 4h^{2}+8h+3 as \left(4h^{2}+2h\right)+\left(6h+3\right).

$$\left(4h^{2}+2h\right)+\left(6h+3\right)$$

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

$$2h\left(2h+1\right)+3\left(2h+1\right)$$

Factor out common term 2h+1 by using distributive property.

$$\left(2h+1\right)\left(2h+3\right)$$

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.

$$4h^{2}+8h+3=0$$

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.

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

Square 8.

$$h=\frac{-8±\sqrt{64-4\times 4\times 3}}{2\times 4}$$

Multiply -4 times 4.

$$h=\frac{-8±\sqrt{64-16\times 3}}{2\times 4}$$

Multiply -16 times 3.

$$h=\frac{-8±\sqrt{64-48}}{2\times 4}$$

Add 64 to -48.

$$h=\frac{-8±\sqrt{16}}{2\times 4}$$

Take the square root of 16.

$$h=\frac{-8±4}{2\times 4}$$

Multiply 2 times 4.

$$h=\frac{-8±4}{8}$$

Now solve the equation h=\frac{-8±4}{8} when ± is plus. Add -8 to 4.

$$h=-\frac{4}{8}$$

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

$$h=-\frac{1}{2}$$

Now solve the equation h=\frac{-8±4}{8} when ± is minus. Subtract 4 from -8.

$$h=-\frac{12}{8}$$

Reduce the fraction \frac{-12}{8} to lowest terms by extracting and canceling out 4.

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

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

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

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

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

Add \frac{1}{2} to h by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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

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

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

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

Multiply 2 times 2.

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

Cancel out 4, the greatest common factor in 4 and 4.

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

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 4

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

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

$$r + s = -2 rs = \frac{3}{4}$$

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

$$r = -1 - u s = -1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{4}

$$(-1 - u) (-1 + u) = \frac{3}{4}$$

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

$$1 - u^2 = \frac{3}{4}$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = \frac{3}{4}-1 = -\frac{1}{4}$$

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

$$u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{2}$$

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

$$r =-1 - \frac{1}{2} = -1.500 s = -1 + \frac{1}{2} = -0.500$$

Evaluate

$$\left(2h+1\right)\left(2h+3\right)$$