Solve 4/9p^2+p-1

$$\frac { 4 } { 9 } p ^ { 2 } + p - 1$$

Factor

$$\frac{\left(4p-3\right)\left(p+3\right)}{9}$$

Solution Steps

Factor out \frac{1}{9}.

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

Consider 4p^{2}+9p-9. Factor the expression by grouping. First, the expression needs to be rewritten as 4p^{2}+ap+bp-9. To find a and b, set up a system to be solved.

$$a+b=9 ab=4\left(-9\right)=-36$$

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 -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 9.

$$a=-3 b=12$$

Rewrite 4p^{2}+9p-9 as \left(4p^{2}-3p\right)+\left(12p-9\right).

$$\left(4p^{2}-3p\right)+\left(12p-9\right)$$

Factor out p in the first and 3 in the second group.

$$p\left(4p-3\right)+3\left(4p-3\right)$$

Factor out common term 4p-3 by using distributive property.

$$\left(4p-3\right)\left(p+3\right)$$

Rewrite the complete factored expression.

$$\frac{\left(4p-3\right)\left(p+3\right)}{9}$$

Evaluate

$$\frac{4p^{2}}{9}+p-1$$