Solve P(a)=3a^2+a-2

$$P ( a ) = 3 a ^ { 2 } + a - 2$$

Factor

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

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.

$$p+q=1 pq=3\left(-2\right)=-6$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

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

Calculate the sum for each pair.

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

The solution is the pair that gives sum 1.

$$p=-2 q=3$$

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

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

Factor out a in 3a^{2}-2a.

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

Factor out common term 3a-2 by using distributive property.

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

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.

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

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.

$$a=\frac{-1±\sqrt{1^{2}-4\times 3\left(-2\right)}}{2\times 3}$$

Square 1.

$$a=\frac{-1±\sqrt{1-4\times 3\left(-2\right)}}{2\times 3}$$

Multiply -4 times 3.

$$a=\frac{-1±\sqrt{1-12\left(-2\right)}}{2\times 3}$$

Multiply -12 times -2.

$$a=\frac{-1±\sqrt{1+24}}{2\times 3}$$

Add 1 to 24.

$$a=\frac{-1±\sqrt{25}}{2\times 3}$$

Take the square root of 25.

$$a=\frac{-1±5}{2\times 3}$$

Multiply 2 times 3.

$$a=\frac{-1±5}{6}$$

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

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

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

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

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

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

Divide -6 by 6.

$$a=-1$$

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

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

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

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

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

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

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

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

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

$$x ^ 2 +\frac{1}{3}x -\frac{2}{3} = 0$$

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

$$r + s = -\frac{1}{3} rs = -\frac{2}{3}$$

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

$$r = -\frac{1}{6} - u s = -\frac{1}{6} + u$$

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

$$(-\frac{1}{6} - u) (-\frac{1}{6} + u) = -\frac{2}{3}$$

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

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

Simplify the expression by subtracting \frac{1}{36} on both sides

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

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

$$u^2 = \frac{25}{36} u = \pm\sqrt{\frac{25}{36}} = \pm \frac{5}{6}$$

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

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

Evaluate

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