Factoring Quartics (FP3)

You'll want to spell maths properly.

You'll want to check the math before you try to 'help' people

You'll want to check the math before you try to 'help' people

By the latter option, I presume they didn't expect you to solve the quartic via direct factorisation. Instead, I suspect you're supposed to note:

$\left(e^x-e^{-x}\right)^2 = e^{2x}+e^{-2x}-2$

And let $w=e^x - e^{-x}$, so that the equation reduces to the quadratic:

$w^2 -7w -8=0$

Which you can solve for $w$ and then solve further quadratics from the definition of $w$ for $e^x$.

However, it's worth pointing out that this is 'secretly' rebuilding the hyperbolic identity for $\cosh 2x$ in terms of $\sinh x$ so it pays to know those instead.

Use identities which is very easy

or using exponentials you get what is known as a symmetric polynomial

Usually if some polynomial of degree more than 2 comes up and its factorisation isn't immediately obvious you probably made a wrong turn..

Hmm, the MS just glosses over the factorisation, But what youre saying makes a lot of sense. Thanks for the help!

I find it amusing how the mark scheme awards marks for the correct quartic, but fails to give the correct quartic itself...

By the latter option, I presume they didn't expect you to solve the quartic via direct factorisation. Instead, I suspect you're supposed to note:

$\left(e^x-e^{-x}\right)^2 = e^{2x}+e^{-2x}-2$

And let $w=e^x - e^{-x}$, so that the equation reduces to the quadratic:

$w^2 -7w -8=0$

Which you can solve for $w$ and then solve further quadratics from the definition of $w$ for $e^x$.

However, it's worth pointing out that this is 'secretly' rebuilding the hyperbolic identity for $\cosh 2x$ in terms of $\sinh x$ so it pays to know those instead.

What do you mean?

The second line?

Posted from TSR Mobile

$-7w=-7e^x+7e^{-x}$. So you would have to be $w=e^x+e^{-x}$.