Question

2 x cube - 11 X square + 44 X + 27 if x is equal to 25 upon 3 - 4 iota​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Answer is 2

Step-by-step explanation:

In this question,

f(x) = $2x^{3}\ -\ 11x^{2}\ +\ 44x\ +\ 27$

We need to find f(x) when x  = $\frac{25}{3-4i}$

Therefore x = $\frac{25\times (3+4i)}{(3-4i)\times (3+4i)}$

                 x = $\frac{25\times (3+4i)}{25}$

                x = 3+4i

f(x) = $2x^{3}\ -\ 11x^{2}\ +\ 44x\ +\ 27$

f(3+4i) = $2(3+4i)^{3}\ -\11(3+4i)^{2}\ +\44(3+4i)\ +\27$

Therefore $(3+4i)^{3}\ =\ (3)^{3}\ +\ (4i)^{3}\ +\ 3\times (3)^{2}\times 4i\ +\ 3\times 3\times (4i)^{2}$

$(3+4i)^{3}\ =\ 27\ +\ (-64i)\ +\ 108i\ -\ 144$

$(3+4i)^{3}\ =\ 44i\ -\ 117$

$2(3+4i)^{3}\ =\ 88i\ - 234$

Similarly, $(3+4i)^{2}\ =\ (3)^{2}\ +\ (4i)^{2}\ +\ 2\times 3\times 4i$

$(3+4i)^{2}\ =\ 9\ -\ 16\ +\ 24i$

$(3+4i)^{2}\ =\ -7\ +\ 24i$

$-11(3+4i)^{2}\ =\ 77\ -\ 264i$

Now, $44(3+4i)\ =\ 132\ +\ 176i$

Hence, f(3+4i) = $2(3+4i)^{3}\ -\11(3+4i)^{2}\ +\44(3+4i)\ +\27$

            f(3+4i) = 88i - 234 + 77 -264i +132 +176i +27

            f(3+4i) = 2

Therefore f(3+4i) = 2