You said that (xy) = 16, and (x+y) = 4 .
From the second equation you can get [ x = 4 - y ],
then substitude that for 'x' in the first equation, and
finally, rearrange the first equation to read
<u>x² - 4x + 16 = 0</u>
Don't even try to factor that quadratic equation. Go straight
to the quadratic formula, and the two solutions you find are ...
<em>x = 2 + i 2√3</em>
and
<em>x = 2 - i 2√3</em> .
Those are the two number that do what you want.
There are no <u>real</u> numbers that can do it.
Answer:
A. 6x^3 - 24x^2 + 6x + 36.
Step-by-step explanation:
(2x - 6)(3x^2 - 3x - 6)
= 2x(3x^2 - 3x - 6) - 6(3x^2-3x-6)
= 6x^3 - 6x^2 - 12x - 18x^2 + 18x + 36
Simplifying like terms:
= 6x^3 - 24x^2 + 6x + 36 (answer).
(5x+1) If X=3 then 5(3)+1=16
(2×2y) if Y=2 then 2×(2)2=8
16÷8=2