Solve 5m^2+51m+54

$$5 m ^ { 2 } + 51 m + 54$$

Factor

$$\left(m+9\right)\left(5m+6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5m^{2}+am+bm+54. To find a and b, set up a system to be solved.

$$a+b=51 ab=5\times 54=270$$

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 270.

$$1,270 2,135 3,90 5,54 6,45 9,30 10,27 15,18$$

Calculate the sum for each pair.

$$1+270=271 2+135=137 3+90=93 5+54=59 6+45=51 9+30=39 10+27=37 15+18=33$$

The solution is the pair that gives sum 51.

$$a=6 b=45$$

Rewrite 5m^{2}+51m+54 as \left(5m^{2}+6m\right)+\left(45m+54\right).

$$\left(5m^{2}+6m\right)+\left(45m+54\right)$$

Factor out m in the first and 9 in the second group.

$$m\left(5m+6\right)+9\left(5m+6\right)$$

Factor out common term 5m+6 by using distributive property.

$$\left(5m+6\right)\left(m+9\right)$$

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.

$$5m^{2}+51m+54=0$$

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.

$$m=\frac{-51±\sqrt{51^{2}-4\times 5\times 54}}{2\times 5}$$

Square 51.

$$m=\frac{-51±\sqrt{2601-4\times 5\times 54}}{2\times 5}$$

Multiply -4 times 5.

$$m=\frac{-51±\sqrt{2601-20\times 54}}{2\times 5}$$

Multiply -20 times 54.

$$m=\frac{-51±\sqrt{2601-1080}}{2\times 5}$$

Add 2601 to -1080.

$$m=\frac{-51±\sqrt{1521}}{2\times 5}$$

Take the square root of 1521.

$$m=\frac{-51±39}{2\times 5}$$

Multiply 2 times 5.

$$m=\frac{-51±39}{10}$$

Now solve the equation m=\frac{-51±39}{10} when ± is plus. Add -51 to 39.

$$m=-\frac{12}{10}$$

Reduce the fraction \frac{-12}{10} to lowest terms by extracting and canceling out 2.

$$m=-\frac{6}{5}$$

Now solve the equation m=\frac{-51±39}{10} when ± is minus. Subtract 39 from -51.

$$m=-\frac{90}{10}$$

Divide -90 by 10.

$$m=-9$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{6}{5} for x_{1} and -9 for x_{2}.

$$5m^{2}+51m+54=5\left(m-\left(-\frac{6}{5}\right)\right)\left(m-\left(-9\right)\right)$$

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

$$5m^{2}+51m+54=5\left(m+\frac{6}{5}\right)\left(m+9\right)$$

Add \frac{6}{5} to m by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$5m^{2}+51m+54=5\times \frac{5m+6}{5}\left(m+9\right)$$

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

$$5m^{2}+51m+54=\left(5m+6\right)\left(m+9\right)$$

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

$$x ^ 2 +\frac{51}{5}x +\frac{54}{5} = 0$$

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

$$r + s = -\frac{51}{5} rs = \frac{54}{5}$$

Two numbers r and s sum up to -\frac{51}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{51}{5} = -\frac{51}{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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{51}{10} - u s = -\frac{51}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{54}{5}

$$(-\frac{51}{10} - u) (-\frac{51}{10} + u) = \frac{54}{5}$$

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

$$\frac{2601}{100} - u^2 = \frac{54}{5}$$

Simplify the expression by subtracting \frac{2601}{100} on both sides

$$-u^2 = \frac{54}{5}-\frac{2601}{100} = -\frac{1521}{100}$$

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

$$u^2 = \frac{1521}{100} u = \pm\sqrt{\frac{1521}{100}} = \pm \frac{39}{10}$$

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

$$r =-\frac{51}{10} - \frac{39}{10} = -9 s = -\frac{51}{10} + \frac{39}{10} = -1.200$$

Evaluate

$$\left(m+9\right)\left(5m+6\right)$$