x (x + 4) = 5
x² + 4x = 5
x²+ x + (-5) = 0
(x+3)²+(t+3)²=9
x² + 6x + 9 + t² + 6t + 9 = 9
x² + t² + 6x + 6t + 18 + (-9) = 0
x² + t² + 6x + 6t + 9 = 0
x² + t² + 6(x + t) + 9 = 0
3/x + 4/2x = x - 1
6/2x + 4/2x = x - 1
10/2x = x - 1
10 = (x - 1) 2x
10 = 2x² - 2x
0 = 2x² + (-2x) + (-10)
a^2m+4m/a^2+4 reduced = m
If you meant :
then this fraction reduced = ![\frac{ma^2+4m+4a^2}{a^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bma%5E2%2B4m%2B4a%5E2%7D%7Ba%5E2%7D)
Hope this helps,
Davinia.
R; all quadratic functions are going to have *ALL REAL NUMBERS*.
Answer:
f(x) = (x+2)³(x²−7x+3)⁴
Step-by-step explanation:
The fundamental theorem of algebra (in its simplest definition), tells us that a polynomial with a degree of n will have n number of roots.
recall that the degree of a polynomial is the highest power that exists in any variable.
i.e.
polynomial, p(x) = axⁿ + bxⁿ⁻¹ + cxⁿ⁻² + .........+ k
has the degree (i.e highest power on a variable x) of n and hence has n-roots
In our case, if we expand all the polynomial choices presented, if we consider the 2nd choice:
f(x) = (x+2)³(x²−7x+3)⁴ (if we expand and simplify this, we end up with)
f(x) = x¹¹−22x¹⁰+150x⁹−116x⁸−2077x⁷+3402x⁶+11158x⁵−8944x⁴−10383x³+13446x²−5076x+648
we notice that the term with the highest power is x¹¹
hence the polynomial has a degree of 11 and hence we expect it to have exactly 11 roots
The incenter, located at the intersection of the angle bisectors.