Question

Find bounds on the real zeros of each polynomial function. $$ f(x)=x^{4}-5 x^{2}-36 $$

Short answer

The bounds on the real zeros are \( -3 \) and \( 3 \).

Step-by-step solution

Identify the Polynomial

Observe the polynomial function: \[ f(x) = x^4 - 5x^2 - 36 \]

Transform into a Quadratic Form

Substitute \(u = x^2\). Then, we get: \[ f(u) = u^2 - 5u - 36 \]

Find the Roots of the Quadratic Equation

Use the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -5\), and \(c = -36\). Substitute these values: \[ u = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-36)}}{2(1)} \]Simplify to find the roots of the quadratic equation:

Calculate the Discriminant

Calculate the discriminant: \[ (-5)^2 - 4(1)(-36) = 25 + 144 = 169 \]

Solve for Roots of the Quadratic Equation

Substituting the discriminant back into the quadratic formula: \[ u = \frac{5 \pm \sqrt{169}}{2} \]We get two solutions: \[ u = \frac{5 + 13}{2} = 9 \]\[ u = \frac{5 - 13}{2} = -4 \]

Re-substitute \(u = x^2\)

Reverse substituting \(u\) to find \(x\): For \(u = 9\): \[ x^2 = 9 \Rightarrow x = \pm3 \]For \(u = -4\): \[ x^2 = -4 \Rightarrow x \text{ is not real} \]

Identify the Real Zeros

The real zeros of the polynomial are: \[ x = \pm3 \]

Determine the Bounds

Since the roots are real and given by \( x = \pm3 \), the bounds for the real zeros are: \[ -3 \] and \[ 3 \]

Key concepts

quadratic formula

The quadratic formula is a powerful tool used to find solutions (or roots) of quadratic equations of the form: \[ ax^2 + bx + c = 0 \]Here, the formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula helps solve quadratic equations by using the coefficients\(a\), \(b\), and \(c\) of the equation.It is derived from completing the square method.In our problem, the polynomial \( f(u) = u^2 - 5u - 36 \) is solved using the quadratic formula to find the values of \(u\). The coefficients for this equation are:

• \( a = 1 \)
• \( b = -5 \)
• \( c = -36 \)

Using the quadratic formula, we substitute these coefficients to find the roots of the quadratic equation.

discriminant

The discriminant is a key part of the quadratic formula. It helps determine the nature of the roots of a quadratic equation.It is found inside the square root of the quadratic formula:\[ b^2 - 4ac \]The value of the discriminant informs us how many real or complex solutions the quadratic equation has:

• If the discriminant is positive (\( > 0 \)), there are two distinct real roots.
• If the discriminant is zero (\( = 0 \)), there is exactly one real root (a repeated root).
• If the discriminant is negative (\(< 0\)), there are two complex roots (no real roots).

In our equation, the discriminant is calculated as:\[ (-5)^2 - 4(1)(-36) = 25 + 144 = 169 \]Since it is positive (169), we know there are two distinct real roots for the quadratic equation.

bounds of zeros

Bounds of zeros refer to the possible range of values within which the real zeros of the polynomial function can lie.Identifying these bounds helps in understanding the behavior of the polynomial and the location of its roots on the number line.For the polynomial \( f(x) = x^4 - 5x^2 - 36 \), after finding the roots \( \pm 3 \), we determine that the real zeros lie between these numbers: \( -3 \) and \( 3 \).Thus, the bounds for the real zeros are \( -3 \) and \( 3 \). This means all real solutions to the equation are within this range.

polynomial function

A polynomial function is an expression consisting of variables, coefficients, and exponents, combined using addition, subtraction, multiplication, and non-negative integer exponents.The general form of a polynomial function is:\[ f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 \]Where:

• \( n \) is a non-negative integer called the degree of the polynomial
• \( a_n, a_{n-1}, ..., a_0 \) are coefficients with \( a_n eq 0 \)
• \( x \) is the variable

The degree of the polynomial dictates its shape and the number of roots it can have.In our case, the polynomial \( f(x) = x^4 - 5x^2 - 36 \) is a fourth-degree polynomial.Using substitution \( u = x^2 \), it transforms into a quadratic polynomial, making it easier to solve for the roots.Then, we convert back to the original variables to locate the real zeros.