Solve 28x^3+19x^2+3x

$$28 x ^ { 3 } + 19 x ^ { 2 } + 3 x$$

Factor

$$x\left(4x+1\right)\left(7x+3\right)$$

Solution Steps

Factor out x.

$$x\left(28x^{2}+19x+3\right)$$

Consider 28x^{2}+19x+3. Factor the expression by grouping. First, the expression needs to be rewritten as 28x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=19 ab=28\times 3=84$$

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 84.

$$1,84 2,42 3,28 4,21 6,14 7,12$$

Calculate the sum for each pair.

$$1+84=85 2+42=44 3+28=31 4+21=25 6+14=20 7+12=19$$

The solution is the pair that gives sum 19.

$$a=7 b=12$$

Rewrite 28x^{2}+19x+3 as \left(28x^{2}+7x\right)+\left(12x+3\right).

$$\left(28x^{2}+7x\right)+\left(12x+3\right)$$

Factor out 7x in the first and 3 in the second group.

$$7x\left(4x+1\right)+3\left(4x+1\right)$$

Factor out common term 4x+1 by using distributive property.

$$\left(4x+1\right)\left(7x+3\right)$$

Rewrite the complete factored expression.

$$x\left(4x+1\right)\left(7x+3\right)$$

Evaluate

$$x\left(4x+1\right)\left(7x+3\right)$$