FInd the roots of this equation: (x^2+6x+8)(x^2-8x+15)=72?

Answer 1

(x^2+6x+8)(x^2-8x+15)=72
{(x+4)(x+2)} {(x-5)(x-3)} = 72
{(x+4)(x-5)} {(x+2)(x-3)} = 72
(x^2-x-20)(x^2-x-6) = 72
put x^2-x = y
(y-20)(y-6)=72
y^2-26y+120-72=0
y^2-26y+48=0
(y-24)(y-2)=0
=> y = 2 or 24
when y=2
x^2-x = 2
=>x^2-x- 2 = 0
=> (x-2)(x+1) = 0
=> x = -1or 2
when y=24
x^2-x = 24
x^2-x- 24=0
=> x = {1+/- sqrt(97)}/2

Answer 2

When you have a degree 4 polynomial equation to solve, you try to get lucky and find a couple of roots by trial and error.

The obvious start here is to write the equation as (x+2)(x+4)(x-3)(x-5) = 72

Can we find a solution were all four of those factors are positive? Well, the largest is 9 more than the smallest, and the first factor 72 has bigger than 9 is 12, and ... well, that won't work. For the same reason, it's unlikely to work when they're all negative. But perhaps we can find something with two positive and two negative. So x would have to be <3 and >-2. +1 doesn't work (there would be a factor of 5). Similarly, -2 doesn't. But +1 and -2 DO work. Hooray!

So multiply out the LHS and subtract 72, and you get f(x) = 0, where f has roots 1 and -2. So f(x) = g(x)(x-1)(x+2). So its other two roots are the roots of g(x), which you can find by the quadratic formula after you find g itself via synthetic division.

Answer 3

Here is a less tedious approach that allows you to find the integral solutions.

(x+4)(x+2) * (x-5)(x-3) = 72
[(x+3)^2 - 1]*[(x-4)^2 - 1] = 72

We could try to find factors of 72 that are both 1 less than a perfect square. 3 and 24 work
so (x+3)^2 = 25 means x = -8 or 2
and (x-4)^2 = 4 means -6 or 2
So x = 2 is an integral solution.

Try also (x+3)^2 = 4 means x = -5 or -1
and (x-4)^2 = 25 means -9 or -1
So x = -1 is also an integral solution.

Answer 4

(x^2 + 6x + 8)(x^2 - 8x + 15) = 72
x^4 - 8x^3 + 15x^2 + 6x^3 - 48x^2...
...+ 90x + 8x^2 - 64x + 120 = 72
x^4 - 2x^3 - 25x^2 + 26x + 48 = 0
(x + 1)(x^3 - 3x^2 - 22x + 48) = 0
(x + 1)(x - 2)(x^2 - x - 24) = 0
x = -1, 2 or (1 +/- √97)/2

Answer 5

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