Using the Factor Theorem to Find Another Factor

Oct 13, 2025 by windailys.com 10 views

Understanding the Problem

The given expression is a quadratic equation in the form of $y^2 - 10y + 24$ . It is stated that this expression has a factor of $y - 4$ . Our goal is to find another factor of the given expression.

Recalling Factor Theorem

To approach this problem, we need to recall the factor theorem, which states that if $f(a) = 0$ , then $(x - a)$ is a factor of the polynomial $f(x)$ . In this case, we are given that $y - 4$ is a factor of the expression $y^2 - 10y + 24$ . This means that when we substitute $y = 4$ into the expression, it should equal zero.

Substituting y = 4 into the Expression

Let's substitute $y = 4$ into the expression $y^2 - 10y + 24$ to verify the given information.

(4)^2 - 10(4) + 24 = 16 - 40 + 24 = 0

As expected, the expression equals zero when $y = 4$ . This confirms that $y - 4$ is indeed a factor of the expression.

Finding Another Factor

Now that we have confirmed the given factor, we need to find another factor of the expression. To do this, we can use polynomial division or synthetic division. However, in this case, we can also use the fact that the product of the two factors must equal the original expression.

Using the Factor Theorem to Find Another Factor

Since we know that $y - 4$ is a factor, we can write the expression as:

y^2 - 10y + 24 = (y - 4)(y - a)

where $a$ is the unknown factor. Our goal is to find the value of $a$ .

Expanding the Product

Let's expand the product $(y - 4)(y - a)$ to get:

(y - 4)(y - a) = y^2 - ay - 4y + 4a

Equating Coefficients

We can equate the coefficients of the expanded product with the original expression:

y^2 - 10y + 24 = y^2 - (a + 4)y + 4a

Comparing Coefficients

Comparing the coefficients of the $y$ term, we get:

-10 = -(a + 4)

Solving for a

Solving for $a$ , we get:

a + 4 = 10

a = 6

Verifying the Result

Now that we have found the value of $a$ , we can verify the result by substituting $a = 6$ into the expression:

(y - 4)(y - 6) = y^2 - 6y - 4y + 24

= y^2 - 10y + 24

As expected, the product of the two factors equals the original expression.

Conclusion

In conclusion, we have found another factor of the expression $y^2 - 10y + 24$ . The other factor is $y - 6$ . This result is verified by substituting $a = 6$ into the expression and confirming that the product of the two factors equals the original expression.

Answer

The correct answer is:

D. $y - 6$

Discussion

This problem requires a good understanding of the factor theorem and polynomial division. It also requires the ability to equate coefficients and solve for unknown variables. The solution involves a combination of algebraic manipulations and logical reasoning.

Practice Problems

Practice problems related to this topic include:

Finding the factors of a quadratic expression
Using the factor theorem to find the factors of a polynomial
Equating coefficients to solve for unknown variables

Conclusion

In conclusion, this problem requires a good understanding of the factor theorem and polynomial division. It also requires the ability to equate coefficients and solve for unknown variables. The solution involves a combination of algebraic manipulations and logical reasoning.

Understanding the Problem

The given expression is a quadratic equation in the form of $y^2 - 10y + 24$ . It is stated that this expression has a factor of $y - 4$ . Our goal is to find another factor of the given expression.

Q: What is the factor theorem?

A: The factor theorem states that if $f(a) = 0$ , then $(x - a)$ is a factor of the polynomial $f(x)$ .

Q: How do we use the factor theorem to find another factor?

A: We can use the factor theorem to find another factor by substituting the given factor into the expression and solving for the unknown variable.

Q: What is the product of the two factors?

A: The product of the two factors is equal to the original expression.

Q: How do we find the value of the unknown variable?

A: We can find the value of the unknown variable by equating the coefficients of the expanded product with the original expression.

Q: What is the value of the unknown variable?

A: The value of the unknown variable is $a = 6$ .

Q: How do we verify the result?

A: We can verify the result by substituting $a = 6$ into the expression and confirming that the product of the two factors equals the original expression.

Q: What is the other factor of the expression?

A: The other factor of the expression is $y - 6$ .

Q: What is the correct answer?

A: The correct answer is:

D. $y - 6$

Q: What are some related topics?

A: Some related topics include:

Factor theorem
Polynomial division
Equating coefficients
Solving for unknown variables

Q: What are some practice problems?

A: Some practice problems include:

Finding the factors of a quadratic expression
Using the factor theorem to find the factors of a polynomial
Equating coefficients to solve for unknown variables

Q: What is the conclusion?

A: In conclusion, this problem requires a good understanding of the factor theorem and polynomial division. It also requires the ability to equate coefficients and solve for unknown variables. The solution involves a combination of algebraic manipulations and logical reasoning.

Frequently Asked Questions

Q: What is the factor theorem?

A: The factor theorem states that if $f(a) = 0$ , then $(x - a)$ is a factor of the polynomial $f(x)$ .

Q: How do I find the factors of a quadratic expression?

A: You can find the factors of a quadratic expression by using the factor theorem and polynomial division.

Q: What is polynomial division?

A: Polynomial division is a method of dividing one polynomial by another to find the quotient and remainder.

Q: How do I equate coefficients?

A: You can equate coefficients by comparing the coefficients of the expanded product with the original expression.

Q: What is the value of the unknown variable?

A: The value of the unknown variable is $a = 6$ .

Q: How do I verify the result?

A: You can verify the result by substituting $a = 6$ into the expression and confirming that the product of the two factors equals the original expression.