Solve 2q^2+5q+3

$$2 q ^ { 2 } + 5 q + 3$$

Factor

$$\left(q+1\right)\left(2q+3\right)$$

Solution Steps

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

$$a+b=5 ab=2\times 3=6$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. 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.

$$a=2 b=3$$

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

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

Factor out 2q in the first and 3 in the second group.

$$2q\left(q+1\right)+3\left(q+1\right)$$

Factor out common term q+1 by using distributive property.

$$\left(q+1\right)\left(2q+3\right)$$

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.

$$2q^{2}+5q+3=0$$

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.

$$q=\frac{-5±\sqrt{5^{2}-4\times 2\times 3}}{2\times 2}$$

Square 5.

$$q=\frac{-5±\sqrt{25-4\times 2\times 3}}{2\times 2}$$

Multiply -4 times 2.

$$q=\frac{-5±\sqrt{25-8\times 3}}{2\times 2}$$

Multiply -8 times 3.

$$q=\frac{-5±\sqrt{25-24}}{2\times 2}$$

Add 25 to -24.

$$q=\frac{-5±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

$$q=\frac{-5±1}{2\times 2}$$

Multiply 2 times 2.

$$q=\frac{-5±1}{4}$$

Now solve the equation q=\frac{-5±1}{4} when ± is plus. Add -5 to 1.

$$q=-\frac{4}{4}$$

Divide -4 by 4.

$$q=-1$$

Now solve the equation q=\frac{-5±1}{4} when ± is minus. Subtract 1 from -5.

$$q=-\frac{6}{4}$$

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

$$q=-\frac{3}{2}$$

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{3}{2} for x_{2}.

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

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

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

Add \frac{3}{2} to q by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Cancel out 2, the greatest common factor in 2 and 2.

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

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 2

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

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

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

Two numbers r and s sum up to -\frac{5}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{2} = -\frac{5}{4}. 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>

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

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

$$(-\frac{5}{4} - u) (-\frac{5}{4} + u) = \frac{3}{2}$$

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

$$\frac{25}{16} - u^2 = \frac{3}{2}$$

Simplify the expression by subtracting \frac{25}{16} on both sides

$$-u^2 = \frac{3}{2}-\frac{25}{16} = -\frac{1}{16}$$

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

$$u^2 = \frac{1}{16} u = \pm\sqrt{\frac{1}{16}} = \pm \frac{1}{4}$$

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

$$r =-\frac{5}{4} - \frac{1}{4} = -1.500 s = -\frac{5}{4} + \frac{1}{4} = -1$$

Evaluate

$$\left(q+1\right)\left(2q+3\right)$$