Solve 5h^2+9h+4

$$5 h ^ { 2 } + 9 h + 4$$

Factor

$$\left(h+1\right)\left(5h+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5h^{2}+ah+bh+4. To find a and b, set up a system to be solved.

$$a+b=9 ab=5\times 4=20$$

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 20.

$$1,20 2,10 4,5$$

Calculate the sum for each pair.

$$1+20=21 2+10=12 4+5=9$$

The solution is the pair that gives sum 9.

$$a=4 b=5$$

Rewrite 5h^{2}+9h+4 as \left(5h^{2}+4h\right)+\left(5h+4\right).

$$\left(5h^{2}+4h\right)+\left(5h+4\right)$$

Factor out h in 5h^{2}+4h.

$$h\left(5h+4\right)+5h+4$$

Factor out common term 5h+4 by using distributive property.

$$\left(5h+4\right)\left(h+1\right)$$

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.

$$5h^{2}+9h+4=0$$

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.

$$h=\frac{-9±\sqrt{9^{2}-4\times 5\times 4}}{2\times 5}$$

Square 9.

$$h=\frac{-9±\sqrt{81-4\times 5\times 4}}{2\times 5}$$

Multiply -4 times 5.

$$h=\frac{-9±\sqrt{81-20\times 4}}{2\times 5}$$

Multiply -20 times 4.

$$h=\frac{-9±\sqrt{81-80}}{2\times 5}$$

Add 81 to -80.

$$h=\frac{-9±\sqrt{1}}{2\times 5}$$

Take the square root of 1.

$$h=\frac{-9±1}{2\times 5}$$

Multiply 2 times 5.

$$h=\frac{-9±1}{10}$$

Now solve the equation h=\frac{-9±1}{10} when ± is plus. Add -9 to 1.

$$h=-\frac{8}{10}$$

Reduce the fraction \frac{-8}{10} to lowest terms by extracting and canceling out 2.

$$h=-\frac{4}{5}$$

Now solve the equation h=\frac{-9±1}{10} when ± is minus. Subtract 1 from -9.

$$h=-\frac{10}{10}$$

Divide -10 by 10.

$$h=-1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{4}{5} for x_{1} and -1 for x_{2}.

$$5h^{2}+9h+4=5\left(h-\left(-\frac{4}{5}\right)\right)\left(h-\left(-1\right)\right)$$

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

$$5h^{2}+9h+4=5\left(h+\frac{4}{5}\right)\left(h+1\right)$$

Add \frac{4}{5} to h by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$5h^{2}+9h+4=5\times \frac{5h+4}{5}\left(h+1\right)$$

Cancel out 5, the greatest common factor in 5 and 5.

$$5h^{2}+9h+4=\left(5h+4\right)\left(h+1\right)$$

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 5

$$x ^ 2 +\frac{9}{5}x +\frac{4}{5} = 0$$

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

$$r + s = -\frac{9}{5} rs = \frac{4}{5}$$

Two numbers r and s sum up to -\frac{9}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{9}{5} = -\frac{9}{10}. 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>

$$r = -\frac{9}{10} - u s = -\frac{9}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{4}{5}

$$(-\frac{9}{10} - u) (-\frac{9}{10} + u) = \frac{4}{5}$$

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

$$\frac{81}{100} - u^2 = \frac{4}{5}$$

Simplify the expression by subtracting \frac{81}{100} on both sides

$$-u^2 = \frac{4}{5}-\frac{81}{100} = -\frac{1}{100}$$

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

$$u^2 = \frac{1}{100} u = \pm\sqrt{\frac{1}{100}} = \pm \frac{1}{10}$$

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

$$r =-\frac{9}{10} - \frac{1}{10} = -1 s = -\frac{9}{10} + \frac{1}{10} = -0.800$$

Evaluate

$$\left(h+1\right)\left(5h+4\right)$$