Solve x-(x+4)^2=x-1

$$x - ( x + 4 ) ^ { 2 } = x - 1$$

Solve for x

$$x=-5$$

$$x=-3$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$x-\left(x^{2}+8x+16\right)=x-1$$

To find the opposite of x^{2}+8x+16, find the opposite of each term.

$$x-x^{2}-8x-16=x-1$$

Combine x and -8x to get -7x.

$$-7x-x^{2}-16=x-1$$

Subtract x from both sides.

$$-7x-x^{2}-16-x=-1$$

Combine -7x and -x to get -8x.

$$-8x-x^{2}-16=-1$$

Add 1 to both sides.

$$-8x-x^{2}-16+1=0$$

Add -16 and 1 to get -15.

$$-8x-x^{2}-15=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}-8x-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=-8 ab=-\left(-15\right)=15$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -8.

$$a=-3 b=-5$$

Rewrite -x^{2}-8x-15 as \left(-x^{2}-3x\right)+\left(-5x-15\right).

$$\left(-x^{2}-3x\right)+\left(-5x-15\right)$$

Factor out x in the first and 5 in the second group.

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

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

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

To find equation solutions, solve -x-3=0 and x+5=0.

$$x=-3 x=-5$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$x-\left(x^{2}+8x+16\right)=x-1$$

To find the opposite of x^{2}+8x+16, find the opposite of each term.

$$x-x^{2}-8x-16=x-1$$

Combine x and -8x to get -7x.

$$-7x-x^{2}-16=x-1$$

Subtract x from both sides.

$$-7x-x^{2}-16-x=-1$$

Combine -7x and -x to get -8x.

$$-8x-x^{2}-16=-1$$

Add 1 to both sides.

$$-8x-x^{2}-16+1=0$$

Add -16 and 1 to get -15.

$$-8x-x^{2}-15=0$$

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.

$$-x^{2}-8x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -8 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-1\right)\left(-15\right)}}{2\left(-1\right)}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-1\right)\left(-15\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-8\right)±\sqrt{64+4\left(-15\right)}}{2\left(-1\right)}$$

Multiply 4 times -15.

$$x=\frac{-\left(-8\right)±\sqrt{64-60}}{2\left(-1\right)}$$

Add 64 to -60.

$$x=\frac{-\left(-8\right)±\sqrt{4}}{2\left(-1\right)}$$

Take the square root of 4.

$$x=\frac{-\left(-8\right)±2}{2\left(-1\right)}$$

The opposite of -8 is 8.

$$x=\frac{8±2}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{8±2}{-2}$$

Now solve the equation x=\frac{8±2}{-2} when ± is plus. Add 8 to 2.

$$x=\frac{10}{-2}$$

Divide 10 by -2.

$$x=-5$$

Now solve the equation x=\frac{8±2}{-2} when ± is minus. Subtract 2 from 8.

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

Divide 6 by -2.

$$x=-3$$

The equation is now solved.

$$x=-5 x=-3$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$x-\left(x^{2}+8x+16\right)=x-1$$

To find the opposite of x^{2}+8x+16, find the opposite of each term.

$$x-x^{2}-8x-16=x-1$$

Combine x and -8x to get -7x.

$$-7x-x^{2}-16=x-1$$

Subtract x from both sides.

$$-7x-x^{2}-16-x=-1$$

Combine -7x and -x to get -8x.

$$-8x-x^{2}-16=-1$$

Add 16 to both sides.

$$-8x-x^{2}=-1+16$$

Add -1 and 16 to get 15.

$$-8x-x^{2}=15$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}-8x=15$$

Divide both sides by -1.

$$\frac{-x^{2}-8x}{-1}=\frac{15}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{8}{-1}\right)x=\frac{15}{-1}$$

Divide -8 by -1.

$$x^{2}+8x=\frac{15}{-1}$$

Divide 15 by -1.

$$x^{2}+8x=-15$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=-15+4^{2}$$

Square 4.

$$x^{2}+8x+16=-15+16$$

Add -15 to 16.

$$x^{2}+8x+16=1$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{1}$$

Simplify.

$$x+4=1 x+4=-1$$

Subtract 4 from both sides of the equation.

$$x=-3 x=-5$$