Let f(x) = (x - a)(x + a)^2 + b, where a and b are positive integers. When f(x) is divided by x - a, the remainder is 5. It is given that f(x) is divisible by x - 2. Find a and b.... Let f(x) = (x - a)(x + a)^2 + b, where a and b are positive integers. When f(x) is divided by x - a, the remainder is 5. It is given that f(x) is divisible by x - 2. Find a and b. Someone claims that the equation f(x) = 0 has no irrational roots. Do you agree? Explain your answer. Let f(x) = ax^3 - 5x^2 + 16x + b, where a and b are constants. It is given that 2x - 1 is a factor of f(x). When f(x) is divided by x - 3, the remainder is 25a. Find a and b. How many integral roots does the equation f(x) = 0 have? Explain your answer.
Understand the Problem
The questions involve polynomial functions, their properties, and factorizations. They require the application of polynomial remainder theorem and root analysis. Each part asks for specific calculations or explanations based on given conditions.
Answer
The cubic polynomial \( p(x) \) can be expressed as \( (7x + 4)(Ax^2 + Bx + 1) \). It has 1 to 3 real roots.
Answer for screen readers
The cubic polynomial ( p(x) ) can be expressed as ( p(x) = (7x + 4)(Ax^2 + Bx + 1) ).
Given ( p(-2) = -6 ), and using the relationships built through factorization and polynomial behavior, we find:
- A degree 3 polynomial has at least 1 and at most 3 real roots. The specific values of ( A ) and ( B ) depend on further information not provided.
Steps to Solve
- Understand the given polynomial properties
We begin with the polynomial ( p(x) ) which is cubic (degree 3). When we divide ( p(x) ) by ( x + 2 ), the remainder is given as -6.
- Apply polynomial remainder theorem
The remainder theorem states that if a polynomial ( f(x) ) is divided by ( x - c ), the remainder is ( f(c) ). Therefore, for division by ( x + 2 ): $$ p(-2) = -6 $$
- Factorization based on the given factor
It is given that ( 7x + 4 ) is a factor of ( p(x) ). This means: $$ p(x) = (7x + 4)(Ax^2 + Bx + C) $$ where ( A, B, C ) are coefficients we need to find.
- Setting up equations
Since the constant term of ( p(x) ) is 4, the constant term in the factorization contributes to this as follows: $$ (7 \cdot 0 + 4)C = 4 \implies C = 1 $$
- Determine ( p(-2) )
Using the equation ( p(x) = (7x + 4)(Ax^2 + Bx + 1) ), we calculate ( p(-2) ): [ p(-2) = (7(-2) + 4)(A(-2)^2 + B(-2) + 1) = -6 ] Calculating gives us: [ p(-2) = (-10)(4A - 2B + 1) = -6 ] Setting this equal to -6 leads us to: [ -10(4A - 2B + 1) = -6 \implies 4A - 2B + 1 = \frac{3}{5} ]
- Finding the number of real roots
To find the number of real roots, we can analyze ( p(x) ) using the discriminant or graphing approach, but since this problem doesn’t provide explicit coefficients for ( p(x) ), we deduce based on the degree and possible behavior of polynomials. A cubic polynomial can have either 1 or 3 real roots.
The cubic polynomial ( p(x) ) can be expressed as ( p(x) = (7x + 4)(Ax^2 + Bx + 1) ).
Given ( p(-2) = -6 ), and using the relationships built through factorization and polynomial behavior, we find:
- A degree 3 polynomial has at least 1 and at most 3 real roots. The specific values of ( A ) and ( B ) depend on further information not provided.
More Information
When dealing with cubic polynomials, their behavior in terms of roots can depend on the coefficients chosen, but they always have at least one real root as per the fundamental theorem of algebra. The remaining roots can either be real or complex.
Tips
- Forgetting to apply the remainder theorem correctly.
- Misidentifying the number of roots based solely on visual inspection without confirming through calculations.
- Not considering polynomial behavior at the endpoints or stationary points.
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