Which of the following polynomials has solution that are real numbers ?
1 answer:
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The second degree polynomial with leading coefficient of -2 and root 4 with multiplicity of 2 is:
<h3>How to write the polynomial?</h3>
A polynomial of degree N, with the N roots {x₁, ..., xₙ} and a leading coefficient a is written as:
Here we know that the degree is 2, the only root is 4 (with a multiplicity of 2, this is equivalent to say that we have two roots at x = 4) and a leading coefficient equal to -2.
Then this polynomial is equal to:
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Answer:
First image attached
The error was done in Step E, because student did not multiply by the negative sign in numerator. Step E must be
.
Second image attached
The error was done in Step C, because the student omitted the of the algebraic identity
. Step C must be
Step-by-step explanation:
First image attached
The error was done in Step E, because student did not multiply by the negative sign in numerator. The real numerator in Step E should be:
Hence, Step E must be .
Second image attached
The error was done in Step C, because the student omitted the of the algebraic identity
. Step C must be
And further steps are described below:
Step D
Which according to the Quadratic Formula, represents a polynomial with complex roots. That is: (,
,
)
(Conjugated complex roots)
Step E
Step F