Solve 3x^2+41x+7

$$3 x ^ { 2 } + 41 x + 7$$

Factor

$$3\left(x-\frac{-\sqrt{1597}-41}{6}\right)\left(x-\frac{\sqrt{1597}-41}{6}\right)$$

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.

$$3x^{2}+41x+7=0$$

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.

$$x=\frac{-41±\sqrt{41^{2}-4\times 3\times 7}}{2\times 3}$$

Square 41.

$$x=\frac{-41±\sqrt{1681-4\times 3\times 7}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-41±\sqrt{1681-12\times 7}}{2\times 3}$$

Multiply -12 times 7.

$$x=\frac{-41±\sqrt{1681-84}}{2\times 3}$$

Add 1681 to -84.

$$x=\frac{-41±\sqrt{1597}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-41±\sqrt{1597}}{6}$$

Now solve the equation x=\frac{-41±\sqrt{1597}}{6} when ± is plus. Add -41 to \sqrt{1597}.

$$x=\frac{\sqrt{1597}-41}{6}$$

Now solve the equation x=\frac{-41±\sqrt{1597}}{6} when ± is minus. Subtract \sqrt{1597} from -41.

$$x=\frac{-\sqrt{1597}-41}{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-41+\sqrt{1597}}{6} for x_{1} and \frac{-41-\sqrt{1597}}{6} for x_{2}.

$$3x^{2}+41x+7=3\left(x-\frac{\sqrt{1597}-41}{6}\right)\left(x-\frac{-\sqrt{1597}-41}{6}\right)$$

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

$$x ^ 2 +\frac{41}{3}x +\frac{7}{3} = 0$$

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

$$r + s = -\frac{41}{3} rs = \frac{7}{3}$$

Two numbers r and s sum up to -\frac{41}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{41}{3} = -\frac{41}{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>

$$r = -\frac{41}{6} - u s = -\frac{41}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{7}{3}

$$(-\frac{41}{6} - u) (-\frac{41}{6} + u) = \frac{7}{3}$$

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

$$\frac{1681}{36} - u^2 = \frac{7}{3}$$

Simplify the expression by subtracting \frac{1681}{36} on both sides

$$-u^2 = \frac{7}{3}-\frac{1681}{36} = -\frac{1597}{36}$$

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

$$u^2 = \frac{1597}{36} u = \pm\sqrt{\frac{1597}{36}} = \pm \frac{\sqrt{1597}}{6}$$

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

$$r =-\frac{41}{6} - \frac{\sqrt{1597}}{6} = -13.494 s = -\frac{41}{6} + \frac{\sqrt{1597}}{6} = -0.173$$

Evaluate

$$3x^{2}+41x+7$$