Complete The Steps To Factor The Polynomial.One Root Of $f(x) = X^3 + X^2 - 22x - 40$ Is 5. If 5 Is A Root Of The Function, Then $\square$ Is A Factor.

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Introduction


Polynomial factorization is a fundamental concept in algebra that involves expressing a polynomial as a product of simpler polynomials. In this article, we will focus on factoring a cubic polynomial using the given root. We will use the factor theorem, which states that if f(a)=0f(a) = 0, then (xβˆ’a)(x - a) is a factor of f(x)f(x). Our goal is to factor the polynomial f(x)=x3+x2βˆ’22xβˆ’40f(x) = x^3 + x^2 - 22x - 40 using the given root 5.

Understanding the Factor Theorem


The factor theorem is a powerful tool in algebra that helps us find the factors of a polynomial. If we know that a polynomial f(x)f(x) has a root aa, then we can write f(x)f(x) as a product of (xβˆ’a)(x - a) and another polynomial g(x)g(x). In other words, f(x)=(xβˆ’a)β‹…g(x)f(x) = (x - a) \cdot g(x). This theorem is a direct consequence of the division algorithm for polynomials.

Applying the Factor Theorem


Given that 5 is a root of the function f(x)=x3+x2βˆ’22xβˆ’40f(x) = x^3 + x^2 - 22x - 40, we can apply the factor theorem to write f(x)f(x) as (xβˆ’5)β‹…g(x)(x - 5) \cdot g(x), where g(x)g(x) is another polynomial. To find g(x)g(x), we can use polynomial long division or synthetic division.

Polynomial Long Division


To divide f(x)=x3+x2βˆ’22xβˆ’40f(x) = x^3 + x^2 - 22x - 40 by (xβˆ’5)(x - 5), we can use polynomial long division. The process involves dividing the leading term of f(x)f(x) by the leading term of (xβˆ’5)(x - 5), which is xx. This gives us x2x^2. We then multiply (xβˆ’5)(x - 5) by x2x^2 and subtract the result from f(x)f(x).

Synthetic Division


Alternatively, we can use synthetic division to divide f(x)f(x) by (xβˆ’5)(x - 5). Synthetic division is a faster and more efficient method than polynomial long division. It involves writing the coefficients of f(x)f(x) in a row and the root 5 below it. We then multiply the root by the first coefficient and add the result to the second coefficient.

Finding the Other Factors


After dividing f(x)f(x) by (xβˆ’5)(x - 5), we get a quadratic polynomial g(x)=x2+6x+8g(x) = x^2 + 6x + 8. To factor g(x)g(x), we can use the quadratic formula or look for two numbers whose product is 8 and whose sum is 6.

Factoring the Quadratic Polynomial


The quadratic polynomial g(x)=x2+6x+8g(x) = x^2 + 6x + 8 can be factored as (x+4)(x+2)(x + 4)(x + 2). Therefore, the complete factorization of f(x)f(x) is (xβˆ’5)(x+4)(x+2)(x - 5)(x + 4)(x + 2).

Conclusion


In this article, we have shown how to factor a cubic polynomial using the given root. We have applied the factor theorem, used polynomial long division and synthetic division, and factored the resulting quadratic polynomial. The complete factorization of f(x)f(x) is (xβˆ’5)(x+4)(x+2)(x - 5)(x + 4)(x + 2). This result can be verified by multiplying the factors together and simplifying the expression.

Example Problems


Problem 1

Factor the polynomial f(x)=x3βˆ’2x2βˆ’11x+12f(x) = x^3 - 2x^2 - 11x + 12 using the given root 3.

Solution

To factor the polynomial f(x)=x3βˆ’2x2βˆ’11x+12f(x) = x^3 - 2x^2 - 11x + 12 using the given root 3, we can apply the factor theorem. We can write f(x)f(x) as (xβˆ’3)β‹…g(x)(x - 3) \cdot g(x), where g(x)g(x) is another polynomial. To find g(x)g(x), we can use polynomial long division or synthetic division.

Problem 2

Factor the polynomial f(x)=x3+2x2βˆ’13xβˆ’10f(x) = x^3 + 2x^2 - 13x - 10 using the given root -2.

Solution

To factor the polynomial f(x)=x3+2x2βˆ’13xβˆ’10f(x) = x^3 + 2x^2 - 13x - 10 using the given root -2, we can apply the factor theorem. We can write f(x)f(x) as (x+2)β‹…g(x)(x + 2) \cdot g(x), where g(x)g(x) is another polynomial. To find g(x)g(x), we can use polynomial long division or synthetic division.

Tips and Tricks


  • When applying the factor theorem, make sure to use the correct root.
  • When using polynomial long division or synthetic division, make sure to follow the correct steps.
  • When factoring a quadratic polynomial, look for two numbers whose product is the constant term and whose sum is the coefficient of the linear term.

Conclusion


In conclusion, factoring polynomials is an essential skill in algebra that involves expressing a polynomial as a product of simpler polynomials. We have shown how to factor a cubic polynomial using the given root, applied the factor theorem, used polynomial long division and synthetic division, and factored the resulting quadratic polynomial. The complete factorization of f(x)f(x) is (xβˆ’5)(x+4)(x+2)(x - 5)(x + 4)(x + 2). This result can be verified by multiplying the factors together and simplifying the expression.

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Introduction


In our previous article, we discussed the steps to factor a cubic polynomial using the given root. We applied the factor theorem, used polynomial long division and synthetic division, and factored the resulting quadratic polynomial. In this article, we will provide a Q&A guide to help you understand the concept of factoring polynomials better.

Q&A


Q1: What is the factor theorem?


A1: The factor theorem is a powerful tool in algebra that helps us find the factors of a polynomial. If we know that a polynomial f(x)f(x) has a root aa, then we can write f(x)f(x) as a product of (xβˆ’a)(x - a) and another polynomial g(x)g(x).

Q2: How do I apply the factor theorem?


A2: To apply the factor theorem, we need to know the root of the polynomial. We can then write the polynomial as (xβˆ’a)β‹…g(x)(x - a) \cdot g(x), where g(x)g(x) is another polynomial. To find g(x)g(x), we can use polynomial long division or synthetic division.

Q3: What is polynomial long division?


A3: Polynomial long division is a method of dividing a polynomial by another polynomial. It involves dividing the leading term of the dividend by the leading term of the divisor, then multiplying the divisor by the result and subtracting the product from the dividend.

Q4: What is synthetic division?


A4: Synthetic division is a faster and more efficient method of dividing a polynomial by another polynomial. It involves writing the coefficients of the dividend in a row and the root below it, then multiplying the root by the first coefficient and adding the result to the second coefficient.

Q5: How do I factor a quadratic polynomial?


A5: To factor a quadratic polynomial, we need to find two numbers whose product is the constant term and whose sum is the coefficient of the linear term. We can then write the quadratic polynomial as the product of two binomials.

Q6: What are the common mistakes to avoid when factoring polynomials?


A6: Some common mistakes to avoid when factoring polynomials include:

  • Not using the correct root when applying the factor theorem
  • Not following the correct steps when using polynomial long division or synthetic division
  • Not looking for two numbers whose product is the constant term and whose sum is the coefficient of the linear term when factoring a quadratic polynomial

Q7: How do I verify the factorization of a polynomial?


A7: To verify the factorization of a polynomial, we can multiply the factors together and simplify the expression. If the result is the original polynomial, then the factorization is correct.

Q8: What are the applications of factoring polynomials?


A8: Factoring polynomials has many applications in algebra and other branches of mathematics. Some of the applications include:

  • Solving systems of equations
  • Finding the roots of a polynomial
  • Factoring quadratic expressions
  • Solving Diophantine equations

Conclusion


In conclusion, factoring polynomials is an essential skill in algebra that involves expressing a polynomial as a product of simpler polynomials. We have provided a Q&A guide to help you understand the concept of factoring polynomials better. By following the steps outlined in this article, you can master the art of factoring polynomials and apply it to solve a wide range of problems in algebra and other branches of mathematics.

Example Problems


Problem 1

Factor the polynomial f(x)=x3+2x2βˆ’13xβˆ’10f(x) = x^3 + 2x^2 - 13x - 10 using the given root -2.

Solution

To factor the polynomial f(x)=x3+2x2βˆ’13xβˆ’10f(x) = x^3 + 2x^2 - 13x - 10 using the given root -2, we can apply the factor theorem. We can write f(x)f(x) as (x+2)β‹…g(x)(x + 2) \cdot g(x), where g(x)g(x) is another polynomial. To find g(x)g(x), we can use polynomial long division or synthetic division.

Problem 2

Factor the polynomial f(x)=x3βˆ’2x2βˆ’11x+12f(x) = x^3 - 2x^2 - 11x + 12 using the given root 3.

Solution

To factor the polynomial f(x)=x3βˆ’2x2βˆ’11x+12f(x) = x^3 - 2x^2 - 11x + 12 using the given root 3, we can apply the factor theorem. We can write f(x)f(x) as (xβˆ’3)β‹…g(x)(x - 3) \cdot g(x), where g(x)g(x) is another polynomial. To find g(x)g(x), we can use polynomial long division or synthetic division.

Tips and Tricks


  • When applying the factor theorem, make sure to use the correct root.
  • When using polynomial long division or synthetic division, make sure to follow the correct steps.
  • When factoring a quadratic polynomial, look for two numbers whose product is the constant term and whose sum is the coefficient of the linear term.