Write the expression (2−3i)+5i(1−3i) as a complex number in standard form.
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We need to use rational root theorem to find out roots here.
The rational root theorem states that if p(x) is a polynomial with integer coefficients and if is a zero of p(x) then p is a factor of constant term and q is a factor of leasing term coefficient.
Here factors of constant term are 1,2,3,4,6,8,12,24,-1,-2,-3,-4,-6,-8,-12, and -24.
And factors of leading coefficient is -1,1.
Hence possible roots may be -1,1,-2,2,-3,3,-4,4,-6,6,-8,8,-12,12,-24 and 24.
Let us plugin these in f(x) to find zeroes.
Hence x=-1 is a zero which means x-(-1)=x+1 is a factor.
Let us use synthetic division to find quotient.
-1 | 1 10 35 50 24
<u>| 0 -1 -9 -26 -24</u>
<u> 1 9 26 24 0</u>
Hence quotient is
Since all coefficients are positive, root must be negative. Let's plugin all remaining negative numbers in the quotient.
Hence x+2 is another factor.
Let us find quotient again using synthetic division.
-2 | 1 9 26 24
<u>| 0 -2 -14 -24</u>
<u> 1 7 12 0</u>
Hence quotient is
Again we got quotient with all positive coefficients, let us plugin remaining negative numbers from rational root theorem.
Hence x+3 is also a factor.
Let us find quotient using synthetic division.
-3 | 1 7 12
<u>| 0 -3 -12</u>
<u> 1 4 0</u>
Hence quotient is x+4.
So,
Please have a look at the graph attached.
