Solve 4x^2+14x+6?

$$4 x ^ { 2 } + 14 x + 6 ?$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(2x^{2}+7x+3\right)$$

Consider 2x^{2}+7x+3. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=7 ab=2\times 3=6$$

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 7.

$$a=1 b=6$$

Rewrite 2x^{2}+7x+3 as \left(2x^{2}+x\right)+\left(6x+3\right).

$$\left(2x^{2}+x\right)+\left(6x+3\right)$$

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

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

Factor out common term 2x+1 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$4x^{2}+14x+6=0$$

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.

$$x=\frac{-14±\sqrt{14^{2}-4\times 4\times 6}}{2\times 4}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\times 4\times 6}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-14±\sqrt{196-16\times 6}}{2\times 4}$$

Multiply -16 times 6.

$$x=\frac{-14±\sqrt{196-96}}{2\times 4}$$

Add 196 to -96.

$$x=\frac{-14±\sqrt{100}}{2\times 4}$$

Take the square root of 100.

$$x=\frac{-14±10}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-14±10}{8}$$

Now solve the equation x=\frac{-14±10}{8} when ± is plus. Add -14 to 10.

$$x=-\frac{4}{8}$$

Reduce the fraction \frac{-4}{8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{1}{2}$$

Now solve the equation x=\frac{-14±10}{8} when ± is minus. Subtract 10 from -14.

$$x=-\frac{24}{8}$$

Divide -24 by 8.

$$x=-3$$

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

$$4x^{2}+14x+6=4\left(x-\left(-\frac{1}{2}\right)\right)\left(x-\left(-3\right)\right)$$

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

$$4x^{2}+14x+6=4\left(x+\frac{1}{2}\right)\left(x+3\right)$$

Add \frac{1}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}+14x+6=4\times \frac{2x+1}{2}\left(x+3\right)$$

Cancel out 2, the greatest common factor in 4 and 2.

$$4x^{2}+14x+6=2\left(2x+1\right)\left(x+3\right)$$

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 4

$$x ^ 2 +\frac{7}{2}x +\frac{3}{2} = 0$$

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

$$r + s = -\frac{7}{2} rs = \frac{3}{2}$$

Two numbers r and s sum up to -\frac{7}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{2} = -\frac{7}{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>

$$r = -\frac{7}{4} - u s = -\frac{7}{4} + u$$

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

$$(-\frac{7}{4} - u) (-\frac{7}{4} + u) = \frac{3}{2}$$

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

$$\frac{49}{16} - u^2 = \frac{3}{2}$$

Simplify the expression by subtracting \frac{49}{16} on both sides

$$-u^2 = \frac{3}{2}-\frac{49}{16} = -\frac{25}{16}$$

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

$$u^2 = \frac{25}{16} u = \pm\sqrt{\frac{25}{16}} = \pm \frac{5}{4}$$

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

$$r =-\frac{7}{4} - \frac{5}{4} = -3 s = -\frac{7}{4} + \frac{5}{4} = -0.500$$

Evaluate

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