Question

How do you solve `m^4-2m^3-35m^2<=0` using a sign chart?

Answer 1

`m^2(m^2-2m-35) <= 0` when `-5 <= m <= 7`

Explanation:

Let us factorize The LHS of `m^4-2m^3-35m^2 <= 0`

i.e. `m^2(m^2-2m-35) <= 0`

or `m^2(m-7)(m+5) <= 0`

Observe that as `m^2 >=0`, the above will satisfy only if

`(m-7)(m+5) < = 0`

Leaving equality for the time being,the product `(m-7)(m+5)` is desired to be negative.

It is apparent that sign of binomials `(m-7)` and `(m+5)` will change around the values `-5` and `7` respectively. In sign chart we divide the real number line using these values, i.e. below `-5`, between `-5` and `7` and above `7` and see how the sign of `(m-7)(m+5)` changes.

Sign Chart

`color(white)(XXXXXXXXXXX)-5color(white)(XXXXX)7`

`(m+5)color(white)(XXX)-ive color(white)(XXXX)+ive color(white)(XXXX)+ive`

`(m-7)color(white)(XXX)-ive color(white)(XXXX)-ive color(white)(XXXX)+ive`

`(m+5)(m-7)color(white)()+ive color(white)(XXX)-ive color(white)(XXXX)+ive`

It is observed that `(m+5)(m-7)<0` when `m` is between `-5` and `7`, which is the solution for the inequality. Wealso addequality sign as we have `m^2(m^2-2m-35) <= 0`

Hence `m^2(m^2-2m-35) <= 0` when `-5 <= m <= 7`