Solve 5a^2-27a+28=

$$5 a ^ { 2 } - 27 a + 28 =$$

Factor

$$\left(a-4\right)\left(5a-7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5a^{2}+pa+qa+28. To find p and q, set up a system to be solved.

$$p+q=-27 pq=5\times 28=140$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. 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.

$$p=-20 q=-7$$

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

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

Factor out 5a in the first and -7 in the second group.

$$5a\left(a-4\right)-7\left(a-4\right)$$

Factor out common term a-4 by using distributive property.

$$\left(a-4\right)\left(5a-7\right)$$

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.

$$5a^{2}-27a+28=0$$

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.

$$a=\frac{-\left(-27\right)±\sqrt{\left(-27\right)^{2}-4\times 5\times 28}}{2\times 5}$$

Square -27.

$$a=\frac{-\left(-27\right)±\sqrt{729-4\times 5\times 28}}{2\times 5}$$

Multiply -4 times 5.

$$a=\frac{-\left(-27\right)±\sqrt{729-20\times 28}}{2\times 5}$$

Multiply -20 times 28.

$$a=\frac{-\left(-27\right)±\sqrt{729-560}}{2\times 5}$$

Add 729 to -560.

$$a=\frac{-\left(-27\right)±\sqrt{169}}{2\times 5}$$

Take the square root of 169.

$$a=\frac{-\left(-27\right)±13}{2\times 5}$$

The opposite of -27 is 27.

$$a=\frac{27±13}{2\times 5}$$

Multiply 2 times 5.

$$a=\frac{27±13}{10}$$

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

$$a=\frac{40}{10}$$

Divide 40 by 10.

$$a=4$$

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

$$a=\frac{14}{10}$$

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

$$a=\frac{7}{5}$$

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

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

Subtract \frac{7}{5} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

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

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

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.

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

Evaluate

$$\left(a-4\right)\left(5a-7\right)$$