Solve 5x^2+27x+28

$$5 x ^ { 2 } + 27 x + 28$$

Factor

$$\left(x+4\right)\left(5x+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx+28. To find a and b, set up a system to be solved.

$$a+b=27 ab=5\times 28=140$$

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 140.

$$1,140 2,70 4,35 5,28 7,20 10,14$$

Calculate the sum for each pair.

$$1+140=141 2+70=72 4+35=39 5+28=33 7+20=27 10+14=24$$

The solution is the pair that gives sum 27.

$$a=7 b=20$$

Rewrite 5x^{2}+27x+28 as \left(5x^{2}+7x\right)+\left(20x+28\right).

$$\left(5x^{2}+7x\right)+\left(20x+28\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(5x+7\right)+4\left(5x+7\right)$$

Factor out common term 5x+7 by using distributive property.

$$\left(5x+7\right)\left(x+4\right)$$

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.

$$5x^{2}+27x+28=0$$

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.

$$x=\frac{-27±\sqrt{27^{2}-4\times 5\times 28}}{2\times 5}$$

Square 27.

$$x=\frac{-27±\sqrt{729-4\times 5\times 28}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-27±\sqrt{729-20\times 28}}{2\times 5}$$

Multiply -20 times 28.

$$x=\frac{-27±\sqrt{729-560}}{2\times 5}$$

Add 729 to -560.

$$x=\frac{-27±\sqrt{169}}{2\times 5}$$

Take the square root of 169.

$$x=\frac{-27±13}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-27±13}{10}$$

Now solve the equation x=\frac{-27±13}{10} when ± is plus. Add -27 to 13.

$$x=-\frac{14}{10}$$

Reduce the fraction \frac{-14}{10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{7}{5}$$

Now solve the equation x=\frac{-27±13}{10} when ± is minus. Subtract 13 from -27.

$$x=-\frac{40}{10}$$

Divide -40 by 10.

$$x=-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{7}{5} for x_{1} and -4 for x_{2}.

$$5x^{2}+27x+28=5\left(x-\left(-\frac{7}{5}\right)\right)\left(x-\left(-4\right)\right)$$

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

$$5x^{2}+27x+28=5\left(x+\frac{7}{5}\right)\left(x+4\right)$$

Add \frac{7}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$5x^{2}+27x+28=5\times \frac{5x+7}{5}\left(x+4\right)$$

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

$$5x^{2}+27x+28=\left(5x+7\right)\left(x+4\right)$$

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

$$x ^ 2 +\frac{27}{5}x +\frac{28}{5} = 0$$

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

$$r + s = -\frac{27}{5} rs = \frac{28}{5}$$

Two numbers r and s sum up to -\frac{27}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{27}{5} = -\frac{27}{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>

$$r = -\frac{27}{10} - u s = -\frac{27}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{28}{5}

$$(-\frac{27}{10} - u) (-\frac{27}{10} + u) = \frac{28}{5}$$

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

$$\frac{729}{100} - u^2 = \frac{28}{5}$$

Simplify the expression by subtracting \frac{729}{100} on both sides

$$-u^2 = \frac{28}{5}-\frac{729}{100} = -\frac{169}{100}$$

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

$$u^2 = \frac{169}{100} u = \pm\sqrt{\frac{169}{100}} = \pm \frac{13}{10}$$

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

$$r =-\frac{27}{10} - \frac{13}{10} = -4.000 s = -\frac{27}{10} + \frac{13}{10} = -1.400$$

Evaluate

$$\left(x+4\right)\left(5x+7\right)$$