One Root Of F ( X ) = X 3 + 10 X 2 − 25 X − 250 F(x)=x^3+10x^2-25x-250 F ( X ) = X 3 + 10 X 2 − 25 X − 250 Is X = − 10 X=-10 X = − 10 . What Are All The Roots Of The Function? Use The Remainder Theorem.A. X = − 25 X=-25 X = − 25 Or X = 10 X=10 X = 10 B. X = − 25 , X = 1 X=-25, X=1 X = − 25 , X = 1 , Or X = 10 X=10 X = 10 C. X = − 10 X=-10 X = − 10 Or

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Introduction

When given a polynomial function and one of its roots, we can use the Remainder Theorem to find the other roots. This theorem states that if a polynomial f(x)f(x) is divided by (xa)(x - a), then the remainder is f(a)f(a). In this case, we are given the cubic function f(x)=x3+10x225x250f(x) = x^3 + 10x^2 - 25x - 250 and one of its roots, x=10x = -10. Our goal is to find all the roots of the function using the Remainder Theorem.

The Remainder Theorem

The Remainder Theorem is a fundamental concept in algebra that helps us find the remainder of a polynomial when divided by another polynomial. In this case, we are interested in finding the remainder when the polynomial f(x)f(x) is divided by (x+10)(x + 10), since we know that x=10x = -10 is a root of the function.

Applying the Remainder Theorem

To apply the Remainder Theorem, we need to evaluate the polynomial f(x)f(x) at x=10x = -10. This will give us the remainder when f(x)f(x) is divided by (x+10)(x + 10).

f(10)=(10)3+10(10)225(10)250f(-10) = (-10)^3 + 10(-10)^2 - 25(-10) - 250

f(10)=1000+1000+250250f(-10) = -1000 + 1000 + 250 - 250

f(10)=0f(-10) = 0

Since the remainder is 0, we know that (x+10)(x + 10) is a factor of the polynomial f(x)f(x). This means that x+10x + 10 is a linear factor of the polynomial, and we can write f(x)f(x) as:

f(x)=(x+10)(x2+ax+b)f(x) = (x + 10)(x^2 + ax + b)

where aa and bb are constants.

Finding the Quadratic Factor

To find the quadratic factor, we need to divide the polynomial f(x)f(x) by (x+10)(x + 10). This will give us the quadratic factor, which we can then use to find the other roots of the function.

Dividing the Polynomial

To divide the polynomial f(x)f(x) by (x+10)(x + 10), we can use long division or synthetic division. Let's use synthetic division to find the quadratic factor.

1 10 -25 -250
-10 -100 250 0

The result of the synthetic division is:

x2+(1010)x+(25+100)=x2+75x^2 + (10 - 10)x + (-25 + 100) = x^2 + 75

So, the quadratic factor is x2+75x^2 + 75.

Finding the Other Roots

Now that we have the quadratic factor, we can find the other roots of the function. To do this, we need to find the roots of the quadratic equation x2+75=0x^2 + 75 = 0.

Solving the Quadratic Equation

To solve the quadratic equation x2+75=0x^2 + 75 = 0, we can use the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In this case, a=1a = 1, b=0b = 0, and c=75c = 75. Plugging these values into the quadratic formula, we get:

x = \{0 \pm \sqrt{0^2 - 4(1)(75)}}{2(1)}

x=0±3002x = \frac{0 \pm \sqrt{-300}}{2}

x=0±10i32x = \frac{0 \pm 10i\sqrt{3}}{2}

x=5i3x = 5i\sqrt{3} or x=5i3x = -5i\sqrt{3}

Conclusion

In conclusion, we have used the Remainder Theorem to find all the roots of the cubic function f(x)=x3+10x225x250f(x) = x^3 + 10x^2 - 25x - 250. We first found that x=10x = -10 is a root of the function, and then used synthetic division to find the quadratic factor x2+75x^2 + 75. Finally, we solved the quadratic equation x2+75=0x^2 + 75 = 0 to find the other roots of the function, which are x=5i3x = 5i\sqrt{3} and x=5i3x = -5i\sqrt{3}.

Final Answer

The final answer is: B\boxed{B}

Introduction

In our previous article, we used the Remainder Theorem to find all the roots of the cubic function f(x)=x3+10x225x250f(x) = x^3 + 10x^2 - 25x - 250. We first found that x=10x = -10 is a root of the function, and then used synthetic division to find the quadratic factor x2+75x^2 + 75. Finally, we solved the quadratic equation x2+75=0x^2 + 75 = 0 to find the other roots of the function, which are x=5i3x = 5i\sqrt{3} and x=5i3x = -5i\sqrt{3}.

Q&A

Q1: What is the Remainder Theorem?

A1: The Remainder Theorem is a fundamental concept in algebra that helps us find the remainder of a polynomial when divided by another polynomial. In this case, we are interested in finding the remainder when the polynomial f(x)f(x) is divided by (x+10)(x + 10), since we know that x=10x = -10 is a root of the function.

Q2: How do we apply the Remainder Theorem?

A2: To apply the Remainder Theorem, we need to evaluate the polynomial f(x)f(x) at x=10x = -10. This will give us the remainder when f(x)f(x) is divided by (x+10)(x + 10).

Q3: What is the quadratic factor of the polynomial f(x)f(x)?

A3: The quadratic factor of the polynomial f(x)f(x) is x2+75x^2 + 75.

Q4: How do we find the other roots of the function?

A4: To find the other roots of the function, we need to find the roots of the quadratic equation x2+75=0x^2 + 75 = 0. We can use the quadratic formula to solve this equation.

Q5: What are the other roots of the function?

A5: The other roots of the function are x=5i3x = 5i\sqrt{3} and x=5i3x = -5i\sqrt{3}.

Q6: Why do we need to use the Remainder Theorem to find the roots of the function?

A6: We need to use the Remainder Theorem to find the roots of the function because it helps us find the remainder when the polynomial f(x)f(x) is divided by (x+10)(x + 10), which is a factor of the polynomial.

Q7: Can we use the Remainder Theorem to find the roots of any polynomial?

A7: Yes, we can use the Remainder Theorem to find the roots of any polynomial. However, we need to know one of the roots of the polynomial in order to apply the theorem.

Q8: What is the significance of the Remainder Theorem in algebra?

A8: The Remainder Theorem is a fundamental concept in algebra that helps us find the remainder of a polynomial when divided by another polynomial. It is used to find the roots of polynomials and is an essential tool in algebraic geometry.

Conclusion

In conclusion, we have used the Remainder Theorem to find all the roots of the cubic function f(x)=x3+10x225x250f(x) = x^3 + 10x^2 - 25x - 250. We have also answered some common questions related to the Remainder Theorem and its application finding the roots of polynomials.

Final Answer

The final answer is: B\boxed{B}