Solving for the polynomial function of least degree with
integral coefficients whose zeros are -5, 3i
We have:
x = -5
Then x + 5 = 0
Therefore one of the factors of the polynomial function is
(x + 5)
Also, we have:
x = 3i
Which can be rewritten as:
x = Sqrt(-9)
Square both sides of the equation:
x^2 = -9
x^2 + 9 = 0
Therefore one of the factors of the polynomial function is (x^2
+ 9)
The polynomial function has factors: (x + 5)(x^2 + 9)
= x(x^2 + 9) + 5(x^2 + 9)
= x^3 + 9x + 5x^2 = 45
Therefore, x^3 + 5x^2 + 9x – 45 = 0
f(x) = x^3 + 5x^2 + 9x – 45
The polynomial function of least degree with integral coefficients
that has the given zeros, -5, 3i is f(x) = x^3 + 5x^2 + 9x – 45
Answer:
(1, 3)
Step-by-step explanation:
Check the picture below.
let's recall that a rhombu's sides are all equal.
<h3> - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - -
</h3>
➷Simply multiply by 2:
3 x 2 = 6
It would be 6 units
➶Hope This Helps You!
➶Good Luck :)
➶Have A Great Day ^-^
↬ Hannah ♡
Answer:
Infinite solutions
Step-by-step explanation:
(x−1)+3=2x+1
(2)(x)+(2)(−1)+3=2x+1(Distribute)
2x+−2+3=2x+1
(2x)+(−2+3)=2x+1(Combine Like Terms)
2x+1=2x+1
2x+1=2x+1
Step 2: Subtract 2x from both sides.
2x+1−2x=2x+1−2x
1=1
Step 3: Subtract 1 from both sides.
1−1=1−1
0=0
