A Quadratic Polynomial Whose Zeroes Are And 4 Is?

If you’re studying math, chances are you’re familiar with the term “quadratic polynomial.” A quadratic polynomial is an equation of degree two in a single variable, typically x. This type of equation can be used to solve a variety of different problems and can even be used to determine the zeroes of the equation. In this blog post, we will explore a specific quadratic polynomial whose zeroes are -2 and 4. We will discuss what this means and how it can be solved using an example problem. By the end, you’ll have a better understanding of quadratic polynomials and their applications.

What is a Quadratic Polynomial?

A quadratic polynomial is a polynomial of degree two. The standard form of a quadratic polynomial is:

Where:

is the leading coefficient,
is the constant term, and
is the variable.

The zeroes of a quadratic polynomial are the values of for which . In other words, the zeroes are the solutions to the equation .

For example, consider the quadratic polynomial . The zeroes of this polynomial are and , since and .

What are the Zeroes of a Quadratic Polynomial?

A quadratic polynomial is a second degree polynomial with three terms. The zeroes of a quadratic polynomial are the x-intercepts of the graph of the polynomial. These zeroes can be found by setting the polynomial equal to zero and solving for x. The zeroes of the quadratic polynomial, P(x), are given by:

P(x) = ax^2 + bx + c

The zeroes of P(x) are given by:

x = -b/2a +/- sqrt((b/2a)^2 – c/a)

For example, consider the quadratic polynomial P(x) = x^2 + 2x + 1. The zeroes of this polynomial are given by:

x = -1 +/- sqrt(1 – 1)

= -1 +/- 0

= -1, 0

How to find a Quadratic Polynomial Whose Zeroes Are And 4 Is?

Assuming you want to find a quadratic polynomial with zeroes at -1 and 4 and a y-intercept of 5, there are a few steps you can take. First, recall that the general form of a quadratic equation is:

$$ax^2 + bx + c = 0$$

So, we know that $a neq 0$ because we’re dealing with a quadratic equation. We also know that the sum of the zeroes must equal the coefficient of $x$, so:

$$-(1) + 4 = b$$

Therefore, $b = 3$. And finally, we know that the product of the zeroes must equal the constant term:

$$-(1)(4) = c$$

Which means $c = -4$. So, our final equation is:

$$ax^2 + bx + c = a(x^2 + 3x – 4) = 0$$

Conclusion

We have seen in this article that a quadratic polynomial whose zeroes are -2 and 4 is (x+2)(x-4). This means that the equation of the polynomial is x^2-2x-8. It also shows how to solve any similar problem with two given zeroes, by finding their product first and then applying the factoring technique to get the expression for the polynomial. We hope this article has provided you with a clear understanding of quadratic equations and how to solve them.