Question

12.Factor: 12x^5y^7 – 8x^4y^9

Answer 1

@Mathbreaker @johnweldon1993

Answer 2

12) \(\large 12x^5y^7 – 8x^4y^9\)

Looks like both terms are divisible by 4 (12/4 = 3 and 8/4 = 2) Looks like there is at least x^4 in both terms...and at least y^7 in each term...so all those can be factored out...

\[\large 4x^4y^7(3x - 2y^2)\]

Make sense?

Answer 3

yes

Answer 4

Now for 13)

\[\huge \frac{12x^6 – 8x^5 + 4x }{4x}\]

Well...this can be broken up into

\[\large \frac{12x^6}{4x} - \frac{8x^5}{4x} + \frac{4x}{4x}\]

So do all those calculations

Answer 5

ok, hold on

Answer 6

12x^6-8x^5+x^2

Answer 7

Not quite...

So 12x^6/4x would be 3x^5

You can think about it like this...

12/4 = 3

x^6 / x would be taking away 1 x...so it would become x^5

and so on..