Factorizing exponential equation

Question

I have this equation:

$$2^{2x}−3⋅2^x−10=0$$

Could someone explain how you factorise it to be: $$(2^x+2)(2^x−5)$$

Answer 1

Can you factorize $y^2 - 3y - 10 = 0$ into $(y + 2)(y - 5) = 0$?

If so, then substitute $y = 2^x$.

Answer 2

Do you know how to factorise $t^2 - 3t - 10$ ?

You need to find its roots. You can do so by applying the standard formula or checking the divisors of $10$ (that is, $\pm 1, \pm 2, \pm 5, \pm 10$)

Once you have factored that, set $t = 2^x$ to get the result you want

Answer 3

$$2^{2x}−3⋅2^x−10=0$$

$$2^{2x}−5⋅2^x+2\cdot2^x−10=0$$

$$2^x(2^{x}−5)+2(2^x−5)=0$$

$$(2^x+2)(2^{x}−5)=0$$

Answer 4

Use : $2^x=y$ so that the equation become: $$ y^2-3y -10=0 $$ that you can easely factorize.