Solve p+5/p^2+p=1/p^2+p-p-6/p+1

$$\frac { p + 5 } { p ^ { 2 } + p } = \frac { 1 } { p ^ { 2 } + p } - \frac { p - 6 } { p + 1 }$$

Solve for p

$$p=1$$

$$p=4$$

Steps Using Factoring

Variable p cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by p\left(p+1\right), the least common multiple of p^{2}+p,p+1.

$$p+5=1-p\left(p-6\right)$$

Use the distributive property to multiply p by p-6.

$$p+5=1-\left(p^{2}-6p\right)$$

To find the opposite of p^{2}-6p, find the opposite of each term.

$$p+5=1-p^{2}+6p$$

Subtract 1 from both sides.

$$p+5-1=-p^{2}+6p$$

Subtract 1 from 5 to get 4.

$$p+4=-p^{2}+6p$$

Add p^{2} to both sides.

$$p+4+p^{2}=6p$$

Subtract 6p from both sides.

$$p+4+p^{2}-6p=0$$

Combine p and -6p to get -5p.

$$-5p+4+p^{2}=0$$

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

$$p^{2}-5p+4=0$$

To solve the equation, factor p^{2}-5p+4 using formula p^{2}+\left(a+b\right)p+ab=\left(p+a\right)\left(p+b\right). To find a and b, set up a system to be solved.

$$a+b=-5 ab=4$$

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 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$a=-4 b=-1$$

Rewrite factored expression \left(p+a\right)\left(p+b\right) using the obtained values.

$$\left(p-4\right)\left(p-1\right)$$

To find equation solutions, solve p-4=0 and p-1=0.

$$p=4 p=1$$

Steps Using Factoring By Grouping

Variable p cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by p\left(p+1\right), the least common multiple of p^{2}+p,p+1.

$$p+5=1-p\left(p-6\right)$$

Use the distributive property to multiply p by p-6.

$$p+5=1-\left(p^{2}-6p\right)$$

To find the opposite of p^{2}-6p, find the opposite of each term.

$$p+5=1-p^{2}+6p$$

Subtract 1 from both sides.

$$p+5-1=-p^{2}+6p$$

Subtract 1 from 5 to get 4.

$$p+4=-p^{2}+6p$$

Add p^{2} to both sides.

$$p+4+p^{2}=6p$$

Subtract 6p from both sides.

$$p+4+p^{2}-6p=0$$

Combine p and -6p to get -5p.

$$-5p+4+p^{2}=0$$

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

$$p^{2}-5p+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as p^{2}+ap+bp+4. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\times 4=4$$

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 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$a=-4 b=-1$$

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

$$\left(p^{2}-4p\right)+\left(-p+4\right)$$

Factor out p in the first and -1 in the second group.

$$p\left(p-4\right)-\left(p-4\right)$$

Factor out common term p-4 by using distributive property.

$$\left(p-4\right)\left(p-1\right)$$

To find equation solutions, solve p-4=0 and p-1=0.

$$p=4 p=1$$

Steps Using the Quadratic Formula

Variable p cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by p\left(p+1\right), the least common multiple of p^{2}+p,p+1.

$$p+5=1-p\left(p-6\right)$$

Use the distributive property to multiply p by p-6.

$$p+5=1-\left(p^{2}-6p\right)$$

To find the opposite of p^{2}-6p, find the opposite of each term.

$$p+5=1-p^{2}+6p$$

Subtract 1 from both sides.

$$p+5-1=-p^{2}+6p$$

Subtract 1 from 5 to get 4.

$$p+4=-p^{2}+6p$$

Add p^{2} to both sides.

$$p+4+p^{2}=6p$$

Subtract 6p from both sides.

$$p+4+p^{2}-6p=0$$

Combine p and -6p to get -5p.

$$-5p+4+p^{2}=0$$

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.

$$p^{2}-5p+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -5.

$$p=\frac{-\left(-5\right)±\sqrt{25-4\times 4}}{2}$$

Multiply -4 times 4.

$$p=\frac{-\left(-5\right)±\sqrt{25-16}}{2}$$

Add 25 to -16.

$$p=\frac{-\left(-5\right)±\sqrt{9}}{2}$$

Take the square root of 9.

$$p=\frac{-\left(-5\right)±3}{2}$$

The opposite of -5 is 5.

$$p=\frac{5±3}{2}$$

Now solve the equation p=\frac{5±3}{2} when ± is plus. Add 5 to 3.

$$p=\frac{8}{2}$$

Divide 8 by 2.

$$p=4$$

Now solve the equation p=\frac{5±3}{2} when ± is minus. Subtract 3 from 5.

$$p=\frac{2}{2}$$

Divide 2 by 2.

$$p=1$$

The equation is now solved.

$$p=4 p=1$$

Steps for Completing the Square

Variable p cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by p\left(p+1\right), the least common multiple of p^{2}+p,p+1.

$$p+5=1-p\left(p-6\right)$$

Use the distributive property to multiply p by p-6.

$$p+5=1-\left(p^{2}-6p\right)$$

To find the opposite of p^{2}-6p, find the opposite of each term.

$$p+5=1-p^{2}+6p$$

Add p^{2} to both sides.

$$p+5+p^{2}=1+6p$$

Subtract 6p from both sides.

$$p+5+p^{2}-6p=1$$

Combine p and -6p to get -5p.

$$-5p+5+p^{2}=1$$

Subtract 5 from both sides.

$$-5p+p^{2}=1-5$$

Subtract 5 from 1 to get -4.

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

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.

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

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$p^{2}-5p+\frac{25}{4}=-4+\frac{25}{4}$$

Add -4 to \frac{25}{4}.

$$p^{2}-5p+\frac{25}{4}=\frac{9}{4}$$

Factor p^{2}-5p+\frac{25}{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}.

$$\left(p-\frac{5}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-\frac{5}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$p-\frac{5}{2}=\frac{3}{2} p-\frac{5}{2}=-\frac{3}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$p=4 p=1$$