Answer:
Therefore the required polynomial is
M(x)=0.83(x³+4x²+16x+64)
Step-by-step explanation:
Given that M is a polynomial of degree 3.
So, it has three zeros.
Let the polynomial be
M(x) =a(x-p)(x-q)(x-r)
The two zeros of the polynomial are -4 and 4i.
Since 4i is a complex number. Then the conjugate of 4i is also a zero of the polynomial i.e -4i.
Then,
M(x)= a{x-(-4)}(x-4i){x-(-4i)}
=a(x+4)(x-4i)(x+4i)
=a(x+4){x²-(4i)²} [ applying the formula (a+b)(a-b)=a²-b²]
=a(x+4)(x²-16i²)
=a(x+4)(x²+16) [∵i² = -1]
=a(x³+4x²+16x+64)
Again given that M(0)= 53.12 . Putting x=0 in the polynomial
53.12 =a(0+4.0+16.0+64)

=0.83
Therefore the required polynomial is
M(x)=0.83(x³+4x²+16x+64)
Let 2k represent the original even number (such that k is an integer)
then 2(k + 1) is the next consecutive even number.
4(2k) - 16 = 2(k + 1)
8k - 16 = 2k + 2
6k = 18
k = 3
since 2k represents the original number, then 2(3) = 6 is the oroginal number
Answer: 6
Answer:
x=9
Step-by-step explanation:
Step 1: Simplify by both sides of the equation.
80-3x=53
80+-3x=53
-3x+80=53
Step 2: Subtract 80 from both sides.
-3x+80-80=53-80
-3x=-27
Step 3: Divide by both sides by 3.
-3x/-3 = -27/-3
x=9