Mathematics Exam > Mathematics Questions > Let f(x) = x3 +x2 - x + 15 and g(x) = ...
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Let f(x) = x3 +x2 - x + 15 and g(x) = x3 + 2x2 - x + 15. Then, over Q
- a)f is irreducible and g is reducible
- b)f is reducible and g is irreducible
- c)Both f and g are reducible
- d)Both f and g are irreducible
Correct answer is option 'B'. Can you explain this answer?
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Let f(x) = x3 +x2 - x + 15 and g(x) = x3 + 2x2 - x + 15. Then, ...
A singular matrix is a square matrix that does not have a unique solution, meaning its determinant is zero. Matrices with determinants equal to zero do not have multiplicative inverses. In contrast, non-singular matrices have unique solutions and possess multiplicative inverses.
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Let f(x) = x3 +x2 - x + 15 and g(x) = x3 + 2x2 - x + 15. Then, ...
Let f (x) = x3 + x2 - x + 15
Put x = -3
f(x) = -27 + 9 + 3 + 15 = 0
So, f(x) is reducible,
and g(x) = x3 + 2x2 - x + 15 is not reducible
Thus f is reducible and g is irreducible.
Put x = -3
f(x) = -27 + 9 + 3 + 15 = 0
So, f(x) is reducible,
and g(x) = x3 + 2x2 - x + 15 is not reducible
Thus f is reducible and g is irreducible.
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Let f(x) = x3 +x2 - x + 15 and g(x) = x3 + 2x2 - x + 15. Then, ...
Explanation:
To determine whether a polynomial is irreducible or reducible, we need to check if it can be factored into nontrivial polynomials with coefficients in the same field. In this case, we are considering the polynomials over the rational numbers (Q).
f(x) = x^3 - x^2 - x + 15
To check the irreducibility of f(x), we can use the Rational Root Theorem to find any possible rational roots. The Rational Root Theorem states that if a polynomial has a rational root p/q, where p and q are coprime integers, then p must divide the constant term (15) and q must divide the leading coefficient (1).
In this case, the possible rational roots of f(x) are ±1, ±3, ±5, ±15. However, after substituting these values into f(x), we find that none of them are roots of f(x). Therefore, f(x) does not have any rational roots and cannot be factored into nontrivial polynomials with rational coefficients.
Since f(x) does not have any rational roots, it cannot be factored into linear factors. Therefore, it is irreducible over Q.
g(x) = x^3 - 2x^2 - x + 15
To check the reducibility of g(x), we can again use the Rational Root Theorem to find any possible rational roots. The possible rational roots of g(x) are ±1, ±3, ±5, ±15. After substituting these values into g(x), we find that x = -1 is a root of g(x).
Using synthetic division or long division, we can divide g(x) by (x + 1) and obtain the quotient as x^2 - 3x + 15. This quadratic polynomial cannot be factored further over the rational numbers.
Therefore, g(x) can be factored as (x + 1)(x^2 - 3x + 15), which means it is reducible over Q.
Conclusion:
- f(x) is irreducible over Q.
- g(x) is reducible over Q.
Hence, the correct answer is option 'B': f is reducible and g is irreducible.
To determine whether a polynomial is irreducible or reducible, we need to check if it can be factored into nontrivial polynomials with coefficients in the same field. In this case, we are considering the polynomials over the rational numbers (Q).
f(x) = x^3 - x^2 - x + 15
To check the irreducibility of f(x), we can use the Rational Root Theorem to find any possible rational roots. The Rational Root Theorem states that if a polynomial has a rational root p/q, where p and q are coprime integers, then p must divide the constant term (15) and q must divide the leading coefficient (1).
In this case, the possible rational roots of f(x) are ±1, ±3, ±5, ±15. However, after substituting these values into f(x), we find that none of them are roots of f(x). Therefore, f(x) does not have any rational roots and cannot be factored into nontrivial polynomials with rational coefficients.
Since f(x) does not have any rational roots, it cannot be factored into linear factors. Therefore, it is irreducible over Q.
g(x) = x^3 - 2x^2 - x + 15
To check the reducibility of g(x), we can again use the Rational Root Theorem to find any possible rational roots. The possible rational roots of g(x) are ±1, ±3, ±5, ±15. After substituting these values into g(x), we find that x = -1 is a root of g(x).
Using synthetic division or long division, we can divide g(x) by (x + 1) and obtain the quotient as x^2 - 3x + 15. This quadratic polynomial cannot be factored further over the rational numbers.
Therefore, g(x) can be factored as (x + 1)(x^2 - 3x + 15), which means it is reducible over Q.
Conclusion:
- f(x) is irreducible over Q.
- g(x) is reducible over Q.
Hence, the correct answer is option 'B': f is reducible and g is irreducible.
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