Solve 5p^2+3p/1+p=4

$$\frac { 5 p ^ { 2 } + 3 p } { 1 + p } = 4$$

Solve for p

$$p=-\frac{4}{5}=-0.8$$

$$p=1$$

Steps Using Factoring By Grouping

Variable p cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by p+1.

$$5p^{2}+3p=4\left(p+1\right)$$

Use the distributive property to multiply 4 by p+1.

$$5p^{2}+3p=4p+4$$

Subtract 4p from both sides.

$$5p^{2}+3p-4p=4$$

Combine 3p and -4p to get -p.

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

Subtract 4 from both sides.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5p^{2}+ap+bp-4. To find a and b, set up a system to be solved.

$$a+b=-1 ab=5\left(-4\right)=-20$$

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 -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -1.

$$a=-5 b=4$$

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

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

Factor out 5p in the first and 4 in the second group.

$$5p\left(p-1\right)+4\left(p-1\right)$$

Factor out common term p-1 by using distributive property.

$$\left(p-1\right)\left(5p+4\right)$$

To find equation solutions, solve p-1=0 and 5p+4=0.

$$p=1 p=-\frac{4}{5}$$

Steps Using the Quadratic Formula

Variable p cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by p+1.

$$5p^{2}+3p=4\left(p+1\right)$$

Use the distributive property to multiply 4 by p+1.

$$5p^{2}+3p=4p+4$$

Subtract 4p from both sides.

$$5p^{2}+3p-4p=4$$

Combine 3p and -4p to get -p.

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

Subtract 4 from both sides.

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

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

$$p=\frac{-\left(-1\right)±\sqrt{1-4\times 5\left(-4\right)}}{2\times 5}$$

Multiply -4 times 5.

$$p=\frac{-\left(-1\right)±\sqrt{1-20\left(-4\right)}}{2\times 5}$$

Multiply -20 times -4.

$$p=\frac{-\left(-1\right)±\sqrt{1+80}}{2\times 5}$$

Add 1 to 80.

$$p=\frac{-\left(-1\right)±\sqrt{81}}{2\times 5}$$

Take the square root of 81.

$$p=\frac{-\left(-1\right)±9}{2\times 5}$$

The opposite of -1 is 1.

$$p=\frac{1±9}{2\times 5}$$

Multiply 2 times 5.

$$p=\frac{1±9}{10}$$

Now solve the equation p=\frac{1±9}{10} when ± is plus. Add 1 to 9.

$$p=\frac{10}{10}$$

Divide 10 by 10.

$$p=1$$

Now solve the equation p=\frac{1±9}{10} when ± is minus. Subtract 9 from 1.

$$p=-\frac{8}{10}$$

Reduce the fraction \frac{-8}{10} to lowest terms by extracting and canceling out 2.

$$p=-\frac{4}{5}$$

The equation is now solved.

$$p=1 p=-\frac{4}{5}$$

Steps for Completing the Square

Variable p cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by p+1.

$$5p^{2}+3p=4\left(p+1\right)$$

Use the distributive property to multiply 4 by p+1.

$$5p^{2}+3p=4p+4$$

Subtract 4p from both sides.

$$5p^{2}+3p-4p=4$$

Combine 3p and -4p to get -p.

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

Divide both sides by 5.

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

Dividing by 5 undoes the multiplication by 5.

$$p^{2}-\frac{1}{5}p=\frac{4}{5}$$

Divide -\frac{1}{5}, the coefficient of the x term, by 2 to get -\frac{1}{10}. Then add the square of -\frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}-\frac{1}{5}p+\left(-\frac{1}{10}\right)^{2}=\frac{4}{5}+\left(-\frac{1}{10}\right)^{2}$$

Square -\frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$p^{2}-\frac{1}{5}p+\frac{1}{100}=\frac{4}{5}+\frac{1}{100}$$

Add \frac{4}{5} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$p^{2}-\frac{1}{5}p+\frac{1}{100}=\frac{81}{100}$$

Factor p^{2}-\frac{1}{5}p+\frac{1}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p-\frac{1}{10}\right)^{2}=\frac{81}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-\frac{1}{10}\right)^{2}}=\sqrt{\frac{81}{100}}$$

Simplify.

$$p-\frac{1}{10}=\frac{9}{10} p-\frac{1}{10}=-\frac{9}{10}$$

Add \frac{1}{10} to both sides of the equation.

$$p=1 p=-\frac{4}{5}$$