Solve 4n^2-n+(8n+3)

$$4 n ^ { 2 } - n + ( 8 n + 3 )$$

Factor

$$\left(n+1\right)\left(4n+3\right)$$

Solution Steps

Multiply and combine like terms.

$$4n^{2}+7n+3$$

Factor the expression by grouping. First, the expression needs to be rewritten as 4n^{2}+an+bn+3. To find a and b, set up a system to be solved.

$$a+b=7 ab=4\times 3=12$$

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 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=3 b=4$$

Rewrite 4n^{2}+7n+3 as \left(4n^{2}+3n\right)+\left(4n+3\right).

$$\left(4n^{2}+3n\right)+\left(4n+3\right)$$

Factor out n in 4n^{2}+3n.

$$n\left(4n+3\right)+4n+3$$

Factor out common term 4n+3 by using distributive property.

$$\left(4n+3\right)\left(n+1\right)$$

Evaluate

$$\left(n+1\right)\left(4n+3\right)$$

Solution Steps

Combine -n and 8n to get 7n.

$$4n^{2}+7n+3$$