Question

What are the values of `a` and `b` if `ax^4 - bx^3 + 40x^2 + 24x + 36` is a perfect square?

Answer 1

`a=36` and `b=+-24`.

Explanation:

Since the degree `ax^4-bx^3+40x^2+24x+36` is 4, the square root must have degree 2 (one-half of 4). Thus, it must be a trinomial of the form

`alphax^2+betax+gamma`

If we start with this generic trinomial and square it, we get

`(alphax^2+betax+gamma)^2`

`=color(magenta)(alpha^2)x^4+color(magenta)(2alphabeta)x^3+(color(magenta)(alphagamma + beta^2))x^2+color(magenta)(2betagamma)x+color(magenta)(gamma^2)`

If this is to be equal to `ax^4-bx^3+40x^2+24x+36`, then the corresponding coefficients must be equal. In other words:

`a=alpha^2`
`"–"b=2alphabeta`
`40=alphagamma+beta^2`
`24=2betagamma`
`36=gamma^2`

We can set up a system of linear equations to solve this, or we can solve them one by one as follows:

`36=gamma^2` is easy. Square rooting both sides gives `gamma=+-6`.

Let's see what we get when we let `gamma=6` and plug this into `24=2betagamma`:

`24=2beta(6)`
`color(white)0 2=beta`

So far so good; now what about `40=alphagamma+beta^2`?

`40=alpha(6)+(2)^2`
`40=6alpha+4`
`36=6alpha`
`color(white)0 6=alpha`

No problems there. Now that we have values for each `alpha`, `beta`, and `gamma,` we can plug them into the equations for `a` and `b`:

`a=alpha^2`
`a=(6)^2`
`a=36`

and

`"–"b=2alphabeta`
`"–"b=2(6)(2)`
`"–"b=24" "=>" "b="–"24`

So `a=36` and `b=–24` is a valid solution. These values for `a` and `b` make our equation

`36x^4-("–"24)x^3+40x^2+24x+36`

which simplifies to

`36x^4+24x^3+40x^2+24x+36`.

Bonus:

This isn't the only solution! As you may infer from the minus sign in front of `b`, there may be a solution where the cubic term is negative. In fact, this is what we get when we let `gamma=–6`. I'll leave the calculations as an exercise, but if `gamma=–6`, we get `a=36` and `b=24`, giving us a polynomial of

`36x^4-24x^3+40x^2+24x+36`.