Solve 5x^2-20x-105

$$5 x ^ { 2 } - 20 x - 105$$

Factor

$$5\left(x-7\right)\left(x+3\right)$$

Solution Steps

Factor out 5.

$$5\left(x^{2}-4x-21\right)$$

Consider x^{2}-4x-21. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-21. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\left(-21\right)=-21$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -21.

$$1,-21 3,-7$$

Calculate the sum for each pair.

$$1-21=-20 3-7=-4$$

The solution is the pair that gives sum -4.

$$a=-7 b=3$$

Rewrite x^{2}-4x-21 as \left(x^{2}-7x\right)+\left(3x-21\right).

$$\left(x^{2}-7x\right)+\left(3x-21\right)$$

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

$$x\left(x-7\right)+3\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(x+3\right)$$

Rewrite the complete factored expression.

$$5\left(x-7\right)\left(x+3\right)$$

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.

$$5x^{2}-20x-105=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 5\left(-105\right)}}{2\times 5}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 5\left(-105\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-20\right)±\sqrt{400-20\left(-105\right)}}{2\times 5}$$

Multiply -20 times -105.

$$x=\frac{-\left(-20\right)±\sqrt{400+2100}}{2\times 5}$$

Add 400 to 2100.

$$x=\frac{-\left(-20\right)±\sqrt{2500}}{2\times 5}$$

Take the square root of 2500.

$$x=\frac{-\left(-20\right)±50}{2\times 5}$$

The opposite of -20 is 20.

$$x=\frac{20±50}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{20±50}{10}$$

Now solve the equation x=\frac{20±50}{10} when ± is plus. Add 20 to 50.

$$x=\frac{70}{10}$$

Divide 70 by 10.

$$x=7$$

Now solve the equation x=\frac{20±50}{10} when ± is minus. Subtract 50 from 20.

$$x=-\frac{30}{10}$$

Divide -30 by 10.

$$x=-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 7 for x_{1} and -3 for x_{2}.

$$5x^{2}-20x-105=5\left(x-7\right)\left(x-\left(-3\right)\right)$$

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

$$5x^{2}-20x-105=5\left(x-7\right)\left(x+3\right)$$

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 5

$$x ^ 2 -4x -21 = 0$$

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

$$r + s = 4 rs = -21$$

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

$$r = 2 - u s = 2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -21

$$(2 - u) (2 + u) = -21$$

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

$$4 - u^2 = -21$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = -21-4 = -25$$

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

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

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

$$r =2 - 5 = -3 s = 2 + 5 = 7$$

Evaluate

$$5\left(x-7\right)\left(x+3\right)$$