Solve left(x-4right)^2+(x-1)(x+1)=25

$${ \left(x-4 \right) }^{ 2 } +(x-1)(x+1)=25$$

Solve for x

$$x=-1$$

$$x=5$$

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}.

$$x^{2}-8x+16+\left(x-1\right)\left(x+1\right)=25$$

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$x^{2}-8x+16+x^{2}-1=25$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-8x+16-1=25$$

Subtract 1 from 16 to get 15.

$$2x^{2}-8x+15=25$$

Subtract 25 from both sides.

$$2x^{2}-8x+15-25=0$$

Subtract 25 from 15 to get -10.

$$2x^{2}-8x-10=0$$

Divide both sides by 2.

$$x^{2}-4x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\left(-5\right)=-5$$

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. The only such pair is the system solution.

$$a=-5 b=1$$

Rewrite x^{2}-4x-5 as \left(x^{2}-5x\right)+\left(x-5\right).

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

Factor out x in x^{2}-5x.

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

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x+1\right)$$

To find equation solutions, solve x-5=0 and x+1=0.

$$x=5 x=-1$$

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}.

$$x^{2}-8x+16+\left(x-1\right)\left(x+1\right)=25$$

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$x^{2}-8x+16+x^{2}-1=25$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-8x+16-1=25$$

Subtract 1 from 16 to get 15.

$$2x^{2}-8x+15=25$$

Subtract 25 from both sides.

$$2x^{2}-8x+15-25=0$$

Subtract 25 from 15 to get -10.

$$2x^{2}-8x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -8 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\left(-10\right)}}{2\times 2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\left(-10\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-8\left(-10\right)}}{2\times 2}$$

Multiply -8 times -10.

$$x=\frac{-\left(-8\right)±\sqrt{64+80}}{2\times 2}$$

Add 64 to 80.

$$x=\frac{-\left(-8\right)±\sqrt{144}}{2\times 2}$$

Take the square root of 144.

$$x=\frac{-\left(-8\right)±12}{2\times 2}$$

The opposite of -8 is 8.

$$x=\frac{8±12}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{8±12}{4}$$

Now solve the equation x=\frac{8±12}{4} when ± is plus. Add 8 to 12.

$$x=\frac{20}{4}$$

Divide 20 by 4.

$$x=5$$

Now solve the equation x=\frac{8±12}{4} when ± is minus. Subtract 12 from 8.

$$x=-\frac{4}{4}$$

Divide -4 by 4.

$$x=-1$$

The equation is now solved.

$$x=5 x=-1$$

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}.

$$x^{2}-8x+16+\left(x-1\right)\left(x+1\right)=25$$

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$x^{2}-8x+16+x^{2}-1=25$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-8x+16-1=25$$

Subtract 1 from 16 to get 15.

$$2x^{2}-8x+15=25$$

Subtract 15 from both sides.

$$2x^{2}-8x=25-15$$

Subtract 15 from 25 to get 10.

$$2x^{2}-8x=10$$

Divide both sides by 2.

$$\frac{2x^{2}-8x}{2}=\frac{10}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{8}{2}\right)x=\frac{10}{2}$$

Divide -8 by 2.

$$x^{2}-4x=\frac{10}{2}$$

Divide 10 by 2.

$$x^{2}-4x=5$$

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

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

Square -2.

$$x^{2}-4x+4=5+4$$

Add 5 to 4.

$$x^{2}-4x+4=9$$

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-2\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{9}$$

Simplify.

$$x-2=3 x-2=-3$$

Add 2 to both sides of the equation.

$$x=5 x=-1$$