Solve x^2+32x+3

$$x ^ { 2 } + 32 x + 3$$

Factor

$$\left(x-\left(-\sqrt{253}-16\right)\right)\left(x-\left(\sqrt{253}-16\right)\right)$$

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.

$$x^{2}+32x+3=0$$

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.

$$x=\frac{-32±\sqrt{32^{2}-4\times 3}}{2}$$

Square 32.

$$x=\frac{-32±\sqrt{1024-4\times 3}}{2}$$

Multiply -4 times 3.

$$x=\frac{-32±\sqrt{1024-12}}{2}$$

Add 1024 to -12.

$$x=\frac{-32±\sqrt{1012}}{2}$$

Take the square root of 1012.

$$x=\frac{-32±2\sqrt{253}}{2}$$

Now solve the equation x=\frac{-32±2\sqrt{253}}{2} when ± is plus. Add -32 to 2\sqrt{253}.

$$x=\frac{2\sqrt{253}-32}{2}$$

Divide -32+2\sqrt{253} by 2.

$$x=\sqrt{253}-16$$

Now solve the equation x=\frac{-32±2\sqrt{253}}{2} when ± is minus. Subtract 2\sqrt{253} from -32.

$$x=\frac{-2\sqrt{253}-32}{2}$$

Divide -32-2\sqrt{253} by 2.

$$x=-\sqrt{253}-16$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -16+\sqrt{253} for x_{1} and -16-\sqrt{253} for x_{2}.

$$x^{2}+32x+3=\left(x-\left(\sqrt{253}-16\right)\right)\left(x-\left(-\sqrt{253}-16\right)\right)$$

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.

$$x ^ 2 +32x +3 = 0$$

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

$$r + s = -32 rs = 3$$

Two numbers r and s sum up to -32 exactly when the average of the two numbers is \frac{1}{2}*-32 = -16. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -16 - u s = -16 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 3

$$(-16 - u) (-16 + u) = 3$$

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

$$256 - u^2 = 3$$

Simplify the expression by subtracting 256 on both sides

$$-u^2 = 3-256 = -253$$

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

$$u^2 = 253 u = \pm\sqrt{253} = \pm \sqrt{253}$$

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

$$r =-16 - \sqrt{253} = -31.906 s = -16 + \sqrt{253} = -0.094$$

Evaluate

$$x^{2}+32x+3$$