Solve 6x^3+47x^2+36x+7

$$6 x ^ { 3 } + 47 x ^ { 2 } + 36 x + 7$$

Factor

$$\left(x+7\right)\left(2x+1\right)\left(3x+1\right)$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 7 and q divides the leading coefficient 6. One such root is -\frac{1}{2}. Factor the polynomial by dividing it by 2x+1.

$$\left(2x+1\right)\left(3x^{2}+22x+7\right)$$

Consider 3x^{2}+22x+7. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+7. To find a and b, set up a system to be solved.

$$a+b=22 ab=3\times 7=21$$

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 21.

$$1,21 3,7$$

Calculate the sum for each pair.

$$1+21=22 3+7=10$$

The solution is the pair that gives sum 22.

$$a=1 b=21$$

Rewrite 3x^{2}+22x+7 as \left(3x^{2}+x\right)+\left(21x+7\right).

$$\left(3x^{2}+x\right)+\left(21x+7\right)$$

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

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

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

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

Rewrite the complete factored expression.

$$\left(2x+1\right)\left(3x+1\right)\left(x+7\right)$$

Evaluate

$$\left(x+7\right)\left(2x+1\right)\left(3x+1\right)$$