When we know the roots are r and s we can just multiply (x-r)(x-s) and set it to zero to recover the original quadratic equation. We can then scale to get rid of fractions or common factors if we like.


The linear term coefficient is negative the sum of the roots, which in the case of conjugates is (negative) twice the real part, coefficient +2 here. The constant term is the product of the roots, which in the case of conjugates is the squared magnitude, here
.
We got a 2x in the middle so we get
Answer: 2x
1. No, because every prime factor can not be prime factorized again. It could have more factorizations, but at least one of the factors would be a composite number which is the product of at least two of the prime factors.
2. Prime numbers and 1 do not have their prime factorization.
KT is twice the length of TM.
Multiply TM by 2:
TM = 2y +6
Now set KT and TM to equal and solve:
5y +3 = 2y +6
Subtract 3 from each side:
5y = 2y +3
Subtract 2y from each side:
3y = 3
Divide both sides by 3:
y = 3/3
y = 1
The answer is C.