Solve 5a^2-17a+6=

$$5 a ^ { 2 } - 17 a + 6 =$$

Factor

$$\left(a-3\right)\left(5a-2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5a^{2}+pa+qa+6. To find p and q, set up a system to be solved.

$$p+q=-17 pq=5\times 6=30$$

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 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -17.

$$p=-15 q=-2$$

Rewrite 5a^{2}-17a+6 as \left(5a^{2}-15a\right)+\left(-2a+6\right).

$$\left(5a^{2}-15a\right)+\left(-2a+6\right)$$

Factor out 5a in the first and -2 in the second group.

$$5a\left(a-3\right)-2\left(a-3\right)$$

Factor out common term a-3 by using distributive property.

$$\left(a-3\right)\left(5a-2\right)$$

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.

$$5a^{2}-17a+6=0$$

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.

$$a=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 5\times 6}}{2\times 5}$$

Square -17.

$$a=\frac{-\left(-17\right)±\sqrt{289-4\times 5\times 6}}{2\times 5}$$

Multiply -4 times 5.

$$a=\frac{-\left(-17\right)±\sqrt{289-20\times 6}}{2\times 5}$$

Multiply -20 times 6.

$$a=\frac{-\left(-17\right)±\sqrt{289-120}}{2\times 5}$$

Add 289 to -120.

$$a=\frac{-\left(-17\right)±\sqrt{169}}{2\times 5}$$

Take the square root of 169.

$$a=\frac{-\left(-17\right)±13}{2\times 5}$$

The opposite of -17 is 17.

$$a=\frac{17±13}{2\times 5}$$

Multiply 2 times 5.

$$a=\frac{17±13}{10}$$

Now solve the equation a=\frac{17±13}{10} when ± is plus. Add 17 to 13.

$$a=\frac{30}{10}$$

Divide 30 by 10.

$$a=3$$

Now solve the equation a=\frac{17±13}{10} when ± is minus. Subtract 13 from 17.

$$a=\frac{4}{10}$$

Reduce the fraction \frac{4}{10} to lowest terms by extracting and canceling out 2.

$$a=\frac{2}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and \frac{2}{5} for x_{2}.

$$5a^{2}-17a+6=5\left(a-3\right)\left(a-\frac{2}{5}\right)$$

Subtract \frac{2}{5} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$5a^{2}-17a+6=5\left(a-3\right)\times \frac{5a-2}{5}$$

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

$$5a^{2}-17a+6=\left(a-3\right)\left(5a-2\right)$$

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

$$x ^ 2 -\frac{17}{5}x +\frac{6}{5} = 0$$

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

$$r + s = \frac{17}{5} rs = \frac{6}{5}$$

Two numbers r and s sum up to \frac{17}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{17}{5} = \frac{17}{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>

$$r = \frac{17}{10} - u s = \frac{17}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{6}{5}

$$(\frac{17}{10} - u) (\frac{17}{10} + u) = \frac{6}{5}$$

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

$$\frac{289}{100} - u^2 = \frac{6}{5}$$

Simplify the expression by subtracting \frac{289}{100} on both sides

$$-u^2 = \frac{6}{5}-\frac{289}{100} = -\frac{169}{100}$$

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.

$$r =\frac{17}{10} - \frac{13}{10} = 0.400 s = \frac{17}{10} + \frac{13}{10} = 3$$

Evaluate

$$\left(a-3\right)\left(5a-2\right)$$