1. ## Show this inequality

Show that the expression

$(6+y)(4-y)(y+4)(y-2)-65<0$

for all values of y are real numbers .

I tried expanding the whole thing .. but no idea how to show it ..

$(-y^2-2y+24)(y^2+2y-8)-65<0$
$-y^4-2y^3+8y^2-2y^3-4y^2+16y+24y^2+48y-192-65<0$
$-y^4-4y^3+28y^2+64y-257<0$

Wonder what i can from here ..

2. if you plot the expanded function on a graph and show that it is always less than zero then that should work. is this any help??

3. ## Re :

Originally Posted by hmmmm
if you plot the expanded function on a graph and show that it is always less than zero then that should work. is this any help??

THanks for your reply but besides plotting it on a graph to show , is that any other method such as showing it in the equation itself ?

4. not that i can see apart from differentiating to show maximum value that is less than 257 sorry

Show that the expression

$(6+y)(4-y)(y+4)(y-2)-65<0$

for all values of y are real numbers .
What you want to show is that

$(1)\;\;\;-65 < (y+6)(y+4)(y-2)(y-4)$

You can actually show that

$-64 < (y+6)(y+4)(y-2)(y-4)$

or

$0 < (y+6)(y+4)(y-2)(y-4) +64$

Factoring the RHS gives

$0 < (y^2 + 2y - 16)^2$

which is clearly true which makes (1) true.

6. ## Re :

Thanks Danny .

By the way , how did you factorise $0 < (y+6)(y+4)(y-2)(y-4) +65$ to $0 < (y^2 + 2y - 16)^2$ .

Thanks Danny .

By the way , how did you factorise $0 < (y+6)(y+4)(y-2)(y-4) +65$ to $0 < (y^2 + 2y - 16)^2$ .
When I looked at the factors

$(y+6)(y+4)(y-2)(y-4)$ I saw $(y+6)(y-4)(y+4)(y-2)$

and grouping them in pairs I saw a difference of 2 in y

$(y^2+2y-24)(y^2+2y-8)$

then

$(y^2+2y-16-8)(y^2+2y-16+8)$

If we let $u = y^2 +2y -16$ then $(u-8)(u+8) = u^2-64$

so $(u-8)(u+8)+64 = u^2-64+64 = u^2 > 0$ giving

$(y^2 +2y -16) > 0$.

Hope it helps.