Solve x^2+42x-135

$$x ^ { 2 } + 42 x - 135$$

Factor

$$\left(x-3\right)\left(x+45\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-135. To find a and b, set up a system to be solved.

$$a+b=42 ab=1\left(-135\right)=-135$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -135.

$$-1,135 -3,45 -5,27 -9,15$$

Calculate the sum for each pair.

$$-1+135=134 -3+45=42 -5+27=22 -9+15=6$$

The solution is the pair that gives sum 42.

$$a=-3 b=45$$

Rewrite x^{2}+42x-135 as \left(x^{2}-3x\right)+\left(45x-135\right).

$$\left(x^{2}-3x\right)+\left(45x-135\right)$$

Factor out x in the first and 45 in the second group.

$$x\left(x-3\right)+45\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x+45\right)$$

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.

$$x^{2}+42x-135=0$$

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.

$$x=\frac{-42±\sqrt{42^{2}-4\left(-135\right)}}{2}$$

Square 42.

$$x=\frac{-42±\sqrt{1764-4\left(-135\right)}}{2}$$

Multiply -4 times -135.

$$x=\frac{-42±\sqrt{1764+540}}{2}$$

Add 1764 to 540.

$$x=\frac{-42±\sqrt{2304}}{2}$$

Take the square root of 2304.

$$x=\frac{-42±48}{2}$$

Now solve the equation x=\frac{-42±48}{2} when ± is plus. Add -42 to 48.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{-42±48}{2} when ± is minus. Subtract 48 from -42.

$$x=-\frac{90}{2}$$

Divide -90 by 2.

$$x=-45$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and -45 for x_{2}.

$$x^{2}+42x-135=\left(x-3\right)\left(x-\left(-45\right)\right)$$

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

$$x^{2}+42x-135=\left(x-3\right)\left(x+45\right)$$

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.

$$x ^ 2 +42x -135 = 0$$

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

$$r + s = -42 rs = -135$$

Two numbers r and s sum up to -42 exactly when the average of the two numbers is \frac{1}{2}*-42 = -21. 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>

$$r = -21 - u s = -21 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -135

$$(-21 - u) (-21 + u) = -135$$

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

$$441 - u^2 = -135$$

Simplify the expression by subtracting 441 on both sides

$$-u^2 = -135-441 = -576$$

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

$$u^2 = 576 u = \pm\sqrt{576} = \pm 24$$

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

$$r =-21 - 24 = -45 s = -21 + 24 = 3$$

Evaluate

$$\left(x-3\right)\left(x+45\right)$$