Solve 3p^2+3+10p

$$3 p ^ { 2 } + 3 + 10 p$$

Factor

$$\left(p+3\right)\left(3p+1\right)$$

Solution Steps

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

$$3p^{2}+10p+3$$

Factor the expression by grouping. First, the expression needs to be rewritten as 3p^{2}+ap+bp+3. To find a and b, set up a system to be solved.

$$a+b=10 ab=3\times 3=9$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 10.

$$a=1 b=9$$

Rewrite 3p^{2}+10p+3 as \left(3p^{2}+p\right)+\left(9p+3\right).

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

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

$$p\left(3p+1\right)+3\left(3p+1\right)$$

Factor out common term 3p+1 by using distributive property.

$$\left(3p+1\right)\left(p+3\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$3p^{2}+10p+3=0$$

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.

$$p=\frac{-10±\sqrt{10^{2}-4\times 3\times 3}}{2\times 3}$$

Square 10.

$$p=\frac{-10±\sqrt{100-4\times 3\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$p=\frac{-10±\sqrt{100-12\times 3}}{2\times 3}$$

Multiply -12 times 3.

$$p=\frac{-10±\sqrt{100-36}}{2\times 3}$$

Add 100 to -36.

$$p=\frac{-10±\sqrt{64}}{2\times 3}$$

Take the square root of 64.

$$p=\frac{-10±8}{2\times 3}$$

Multiply 2 times 3.

$$p=\frac{-10±8}{6}$$

Now solve the equation p=\frac{-10±8}{6} when ± is plus. Add -10 to 8.

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

Reduce the fraction \frac{-2}{6} to lowest terms by extracting and canceling out 2.

$$p=-\frac{1}{3}$$

Now solve the equation p=\frac{-10±8}{6} when ± is minus. Subtract 8 from -10.

$$p=-\frac{18}{6}$$

Divide -18 by 6.

$$p=-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{3} for x_{1} and -3 for x_{2}.

$$3p^{2}+10p+3=3\left(p-\left(-\frac{1}{3}\right)\right)\left(p-\left(-3\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$3p^{2}+10p+3=3\left(p+\frac{1}{3}\right)\left(p+3\right)$$

Add \frac{1}{3} to p by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3p^{2}+10p+3=3\times \frac{3p+1}{3}\left(p+3\right)$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3p^{2}+10p+3=\left(3p+1\right)\left(p+3\right)$$

Evaluate

$$\left(p+3\right)\left(3p+1\right)$$