Solve 3a^2-5a+2=

$$3 a ^ { 2 } - 5 a + 2 =$$

Factor

$$\left(a-1\right)\left(3a-2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3a^{2}+pa+qa+2. To find p and q, set up a system to be solved.

$$p+q=-5 pq=3\times 2=6$$

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 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$p=-3 q=-2$$

Rewrite 3a^{2}-5a+2 as \left(3a^{2}-3a\right)+\left(-2a+2\right).

$$\left(3a^{2}-3a\right)+\left(-2a+2\right)$$

Factor out 3a in the first and -2 in the second group.

$$3a\left(a-1\right)-2\left(a-1\right)$$

Factor out common term a-1 by using distributive property.

$$\left(a-1\right)\left(3a-2\right)$$

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.

$$3a^{2}-5a+2=0$$

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.

$$a=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 3\times 2}}{2\times 3}$$

Square -5.

$$a=\frac{-\left(-5\right)±\sqrt{25-4\times 3\times 2}}{2\times 3}$$

Multiply -4 times 3.

$$a=\frac{-\left(-5\right)±\sqrt{25-12\times 2}}{2\times 3}$$

Multiply -12 times 2.

$$a=\frac{-\left(-5\right)±\sqrt{25-24}}{2\times 3}$$

Add 25 to -24.

$$a=\frac{-\left(-5\right)±\sqrt{1}}{2\times 3}$$

Take the square root of 1.

$$a=\frac{-\left(-5\right)±1}{2\times 3}$$

The opposite of -5 is 5.

$$a=\frac{5±1}{2\times 3}$$

Multiply 2 times 3.

$$a=\frac{5±1}{6}$$

Now solve the equation a=\frac{5±1}{6} when ± is plus. Add 5 to 1.

$$a=\frac{6}{6}$$

Divide 6 by 6.

$$a=1$$

Now solve the equation a=\frac{5±1}{6} when ± is minus. Subtract 1 from 5.

$$a=\frac{4}{6}$$

Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.

$$a=\frac{2}{3}$$

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

$$3a^{2}-5a+2=3\left(a-1\right)\left(a-\frac{2}{3}\right)$$

Subtract \frac{2}{3} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3a^{2}-5a+2=3\left(a-1\right)\times \frac{3a-2}{3}$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3a^{2}-5a+2=\left(a-1\right)\left(3a-2\right)$$

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 3

$$x ^ 2 -\frac{5}{3}x +\frac{2}{3} = 0$$

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

$$r + s = \frac{5}{3} rs = \frac{2}{3}$$

Two numbers r and s sum up to \frac{5}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{3} = \frac{5}{6}. 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>

$$r = \frac{5}{6} - u s = \frac{5}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{3}

$$(\frac{5}{6} - u) (\frac{5}{6} + u) = \frac{2}{3}$$

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

$$\frac{25}{36} - u^2 = \frac{2}{3}$$

Simplify the expression by subtracting \frac{25}{36} on both sides

$$-u^2 = \frac{2}{3}-\frac{25}{36} = -\frac{1}{36}$$

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

$$u^2 = \frac{1}{36} u = \pm\sqrt{\frac{1}{36}} = \pm \frac{1}{6}$$

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

$$r =\frac{5}{6} - \frac{1}{6} = 0.667 s = \frac{5}{6} + \frac{1}{6} = 1.000$$

Evaluate

$$\left(a-1\right)\left(3a-2\right)$$