Solve 3a^2+10a+3

$$3 a ^ { 2 } + 10 a + 3$$

Factor

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

Solution Steps

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

$$p+q=10 pq=3\times 3=9$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q are both positive. List all such integer pairs that give product 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 10.

$$p=1 q=9$$

Rewrite 3a^{2}+10a+3 as \left(3a^{2}+a\right)+\left(9a+3\right).

$$\left(3a^{2}+a\right)+\left(9a+3\right)$$

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

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

Factor out common term 3a+1 by using distributive property.

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

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.

$$3a^{2}+10a+3=0$$

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.

$$a=\frac{-10±\sqrt{10^{2}-4\times 3\times 3}}{2\times 3}$$

Square 10.

$$a=\frac{-10±\sqrt{100-4\times 3\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$a=\frac{-10±\sqrt{100-12\times 3}}{2\times 3}$$

Multiply -12 times 3.

$$a=\frac{-10±\sqrt{100-36}}{2\times 3}$$

Add 100 to -36.

$$a=\frac{-10±\sqrt{64}}{2\times 3}$$

Take the square root of 64.

$$a=\frac{-10±8}{2\times 3}$$

Multiply 2 times 3.

$$a=\frac{-10±8}{6}$$

Now solve the equation a=\frac{-10±8}{6} when ± is plus. Add -10 to 8.

$$a=-\frac{2}{6}$$

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

$$a=-\frac{1}{3}$$

Now solve the equation a=\frac{-10±8}{6} when ± is minus. Subtract 8 from -10.

$$a=-\frac{18}{6}$$

Divide -18 by 6.

$$a=-3$$

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

$$3a^{2}+10a+3=3\left(a-\left(-\frac{1}{3}\right)\right)\left(a-\left(-3\right)\right)$$

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

$$3a^{2}+10a+3=3\left(a+\frac{1}{3}\right)\left(a+3\right)$$

Add \frac{1}{3} to a by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3a^{2}+10a+3=3\times \frac{3a+1}{3}\left(a+3\right)$$

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

$$3a^{2}+10a+3=\left(3a+1\right)\left(a+3\right)$$

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

$$x ^ 2 +\frac{10}{3}x +1 = 0$$

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

$$r + s = -\frac{10}{3} rs = 1$$

Two numbers r and s sum up to -\frac{10}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{10}{3} = -\frac{5}{3}. 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>

$$r = -\frac{5}{3} - u s = -\frac{5}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 1

$$(-\frac{5}{3} - u) (-\frac{5}{3} + u) = 1$$

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

$$\frac{25}{9} - u^2 = 1$$

Simplify the expression by subtracting \frac{25}{9} on both sides

$$-u^2 = 1-\frac{25}{9} = -\frac{16}{9}$$

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

$$u^2 = \frac{16}{9} u = \pm\sqrt{\frac{16}{9}} = \pm \frac{4}{3}$$

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

$$r =-\frac{5}{3} - \frac{4}{3} = -3 s = -\frac{5}{3} + \frac{4}{3} = -0.333$$

Evaluate

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