Solve 4x^4-21x^3-46x^2+219x+180

$$4 x ^ { 4 } - 21 x ^ { 3 } - 46 x ^ { 2 } + 219 x + 180$$

Factor

$$\left(x-5\right)\left(x-4\right)\left(x+3\right)\left(4x+3\right)$$

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 180 and q divides the leading coefficient 4. One such root is 5. Factor the polynomial by dividing it by x-5.

$$\left(x-5\right)\left(4x^{3}-x^{2}-51x-36\right)$$

Consider 4x^{3}-x^{2}-51x-36. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -36 and q divides the leading coefficient 4. One such root is -3. Factor the polynomial by dividing it by x+3.

$$\left(x+3\right)\left(4x^{2}-13x-12\right)$$

Consider 4x^{2}-13x-12. Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=-13 ab=4\left(-12\right)=-48$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -48.

$$1,-48 2,-24 3,-16 4,-12 6,-8$$

Calculate the sum for each pair.

$$1-48=-47 2-24=-22 3-16=-13 4-12=-8 6-8=-2$$

The solution is the pair that gives sum -13.

$$a=-16 b=3$$

Rewrite 4x^{2}-13x-12 as \left(4x^{2}-16x\right)+\left(3x-12\right).

$$\left(4x^{2}-16x\right)+\left(3x-12\right)$$

Factor out 4x in the first and 3 in the second group.

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

Factor out common term x-4 by using distributive property.

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

Rewrite the complete factored expression.

$$\left(x-5\right)\left(x-4\right)\left(x+3\right)\left(4x+3\right)$$

Evaluate

$$\left(x-5\right)\left(x-4\right)\left(x+3\right)\left(4x+3\right)$$