Z= -12i + 11 what are the real imaginary parts of z
1 answer:
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Answer:
(x-2)(x+3)(x+6)
Step-by-step explanation:
Given the polynomial function p(x)=x^3+7x^2-36
We are to write it as a product of its linear factor
Assuming the value of x that will make the polynomial p(x) to be zero
Let x = 2
P(2) = 2³+7(2)²-36
P(2) = 8+7(4)-36
P(2) = 8+28-36
P(2) = 0
Since p(2) = 0 hence x-2 is one of the linear factors
Also assume x = -3
P(-3) = (-3)³+7(-3)²-36
P(-3) = -27+7(9)-36
P(-3) = -27+63-36
P(-3) = 36-36
P(-3) = 0
Since p(-3) = 0, hence x+3 is also a factor
The two linear pair are (x-2)(x+3)
(x-2)(x+3) = x²+3x-2x-6
(x-2)(x+3) = x²+x-6
To get the third linear function, we will divide x^3+7x^2-36 by x²+x-6 as shown in the attachment.
x^3+7x^2-36/x²+x-6 = x+6
Hence the third linear factor is x+6
x^3+7x^2-36 = (x-2)(x+3)(x+6)
