Question

How do you solve `2x ^ { 4} + 3x ^ { 3} + 80< 28x ^ { 2} + 12x?`

Answer 1

The solution to the inequality is `(-4,-2)uu(2,2.5)` in bracket notation, or { `xepsilonR` |`-4 < x<-2 or 2 < x < 2.5 }` in set builder notation.

Explanation:

First put all the terms on one side of the inequality:
`2x^4+3x^3-28x^2-12x+80<0`
The polynomial on the left side factors, using the factor theorem:
`(x+2)(x-2)(x+4)(2x-5)<0`

Set up a sign chart (y-chart) with a number line along the top. Only the values of x which are the zeros of the polynomial on the left side need to be indicated on the number line.

`color(white)(aaaaaaaaaaaaa)`-4`color(white)(aaaaaaa)`-2`color(white)(aaaaaaa)`2`color(white)(aaaaaaa)`2.5
#___________________________________________________________________________________________________________________ #(x+2)#`color(white)(aaa)-` `color(white)(aa)` `|` `color(white)(aa)-` `color(white)(aa)` `|` `color(white)(aa)+` `color(white)(aa)` `|` `color(white)(aa)+` `color(white)(aa)` `|` `color(white)(aa)+`
`(x-2)` `color(white)(aaa)-` `color(white)(aa)` `|` `color(white)(aa)-` `color(white)(aa)` `|` `color(white)(aa)-` `color(white)(aa)` `|` `color(white)(aa)+` `color(white)(a)` `color(white)(a)` `|` `color(white)(aa)+`
`(x+4)` `color(white)(aaa)-` `color(white)(aa)` `|` `color(white)(aa)+` `color(white)(aa)` `|` `color(white)(aa)+` `color(white)(aa)` `|` `color(white)(aa)+` `color(white)(aa)` `|` `color(white)(aa)+`
`(2x-5)` `color(white)(aa)-` `color(white)(aa)` `|` `color(white)(aa)-` `color(white)(aa)` `|` `color(white)(aa)-` `color(white)(aa)` `|` `color(white)(aa)-` `color(white)(aa)` `|` `color(white)(aa)+`
`__________________________________________________________________________________________________________________`

`y` `color(white)(aaaaaaaa)+` `color(white)(aa)` `|` `color(white)(aa)-` `color(white)(aa)` `|` `color(white)(aa)+` `color(white)(aa)` `|` `color(white)(aa)-` `color(white)(aa)` `|` `color(white)(aa)+`

Since y is the product of the factors in the left column, we mark on the chart where each factor is positive or negative. For example (x+2) equals 0 when x = -2. So for values of x larger than -2, the factor (x+2) will be positive, and for values of x less than -2, the factor (x+2) will be negative.

Then using the rule that an even number of negatives multiplied is positive and an odd number of negatives multiplied is negative, we use the columns to fill in the bottom row to show where y is positive and negative.

For x values between -4 and -2, the value of y will be negative. Also for x values between 2 and 2.5, the value of y will be negative. Since the inequality is < 0, we want to exclude the values of x which make the expression equal to zero. So we exclude the endpoints of the intervals, giving an answer to the question of `-4 < x < -2 or 2 < x < 2.5`