Understanding the Factor Theorem for Polynomials
The question asks us to determine which of the given polynomials has \((x+2)\) as a factor. We can use the Factor Theorem to solve this efficiently. The Factor Theorem states that for a polynomial \(P(x)\), \((x-a)\) is a factor of \(P(x)\) if and only if \(P(a) = 0\). In this problem, our potential factor is \((x+2)\), which can be written as \((x - (-2))\). Therefore, we need to find the polynomial \(P(x)\) for which \(P(-2) = 0\).
Evaluating Polynomials at \(x = -2\)
We will substitute \(x = -2\) into each of the given polynomial options to check if the result is zero.
Option 1: \(P(x) = x^5-4x^4-3x^3 + 8x^2 - 14x + 12\)
Let's calculate \(P(-2)\): \(P(-2) = (-2)^5 - 4(-2)^4 - 3(-2)^3 + 8(-2)^2 - 14(-2) + 12\) \(P(-2) = (-32) - 4(16) - 3(-8) + 8(4) - (-28) + 12\) \(P(-2) = -32 - 64 + 24 + 32 + 28 + 12\) Combine the negative terms: \(-32 - 64 = -96\) Combine the positive terms: \(24 + 32 + 28 + 12 = 96\) \(P(-2) = -96 + 96 = 0\) Since \(P(-2) = 0\), \((x+2)\) is a factor of this polynomial.
Option 2: \(P(x) = x^5 +4x^4-3x^3 + 8x^2 - 14x + 12\)
Let's calculate \(P(-2)\): \(P(-2) = (-2)^5 + 4(-2)^4 - 3(-2)^3 + 8(-2)^2 - 14(-2) + 12\) \(P(-2) = (-32) + 4(16) - 3(-8) + 8(4) - (-28) + 12\) \(P(-2) = -32 + 64 + 24 + 32 + 28 + 12\) Combine the negative terms: \(-32\) Combine the positive terms: \(64 + 24 + 32 + 28 + 12 = 160\) \(P(-2) = -32 + 160 = 128\) Since \(P(-2) \neq 0\), \((x+2)\) is not a factor of this polynomial.
Option 3: \(P(x) = x^5-4x^4 + 3x^3 + 8x^2 - 14x + 12\)
Let's calculate \(P(-2)\): \(P(-2) = (-2)^5 - 4(-2)^4 + 3(-2)^3 + 8(-2)^2 - 14(-2) + 12\) \(P(-2) = (-32) - 4(16) + 3(-8) + 8(4) - (-28) + 12\) \(P(-2) = -32 - 64 - 24 + 32 + 28 + 12\) Combine the negative terms: \(-32 - 64 - 24 = -120\) Combine the positive terms: \(32 + 28 + 12 = 72\) \(P(-2) = -120 + 72 = -48\) Since \(P(-2) \neq 0\), \((x+2)\) is not a factor of this polynomial.
Option 4: \(P(x) = x^5-4x^4-3x^3 + 8x^2 + 14x + 12\)
Let's calculate \(P(-2)\): \(P(-2) = (-2)^5 - 4(-2)^4 - 3(-2)^3 + 8(-2)^2 + 14(-2) + 12\) \(P(-2) = (-32) - 4(16) - 3(-8) + 8(4) + (-28) + 12\) \(P(-2) = -32 - 64 + 24 + 32 - 28 + 12\) Combine the negative terms: \(-32 - 64 - 28 = -124\) Combine the positive terms: \(24 + 32 + 12 = 68\) \(P(-2) = -124 + 68 = -56\) Since \(P(-2) \neq 0\), \((x+2)\) is not a factor of this polynomial.
Conclusion
Based on the Factor Theorem and our calculations, only the first polynomial, \(x^5-4x^4-3x^3 + 8x^2 - 14x + 12\), results in \(P(-2) = 0\). Therefore, \((x+2)\) is a factor of this polynomial.
Important Questions from Polynomials
In the expansion of (x + 3) 3, the coefficient of x is:
If the degree of polynomial 9 x 5y 2z ris 15, then r = ?
The factorisation of x 2+ 11xy + 24y 2is:
The value of 16x 4+ 25y 2– 40x 2y at x = 5 and y = 2 is:
If the sum of the squares of the zeros of quadratic polynomial f(x) = x 2– 8x + k is 40, then find the value of k.
