Question

Show that the binomial is a factor of the polynomial. Then factor the function completely. $$ t(x)=x^3-5 x^2-9 x+45 ; x-5 $$

Short answer

The binomial \((x-5)\) is indeed a factor of the polynomial \(t(x)=x^3-5x^2-9x+45\). The complete factorization of the polynomial is \((x-5)(x^2-9)\).

Step-by-step solution

Apply Synthetic Division

Set up synthetic division to divide \(t(x)\) by \((x-5)\). Synthetic division is more efficient and less cumbersome to write out than long division of polynomials.

Perform Synthetic Division

To do this, take the coefficients of \(t(x)\) and the value that makes the binomial zero (in this case \(5\)) and set up a table of coefficients. Then add and multiply across the table to get the remaining coefficients. The coefficients of this row now form the coefficients of the quotient.

Write Down The Result

If we get a remainder of zero when we divide then that means the binomial \((x-5)\) is a factor of the polynomial. Now we can write the polynomial \(t(x)\) as the product of the binomial and the said quotient.

Factor The Quotient

To factor the polynomial completely we should also factor the quotient obtained in step 3, if possible. In this case, it is indeed possible to factor the quotient further as it is a quadratic.

Finishing The Exercise

Now replace the quotient in the result from step 3 with its factors. This gives us the completely factored form of the polynomial \(t(x)\).

Key concepts

Synthetic Division

Synthetic division is a simplified way to divide polynomials, especially when dividing by a linear binomial like \((x - a)\). It reduces the complexity and the room for errors that comes with long polynomial division. Here's how it works:

• First, write down the coefficients of the polynomial you want to divide.
• Next, write the value, \(a\), that would make the divisor \((x-a)\) equal to zero.
• Use this \(a\) in a synthetic division setup: bring down the leading coefficient to the bottom row, multiply it by \(a\), and add it to the next coefficient. Continue this process column by column.

At the end of synthetic division, if your remainder is zero, the binomial is a factor. Otherwise, it isn't. For example, dividing \(x^3-5x^2-9x+45\) by \((x-5)\) involves using 5 (the \(x\) value) and processing the coefficients \([1, -5, -9, 45]\). If done correctly, this method shows whether \((x-5)\) is a factor.

Polynomial Division

Polynomial division is a procedure used to divide one polynomial by another. It's similar to arithmetic division but involves variables with exponents. There are two main methods: long division and synthetic division.Long division of polynomials resembles the long division done with numbers:

• Divide the first term of the dividend by the first term of the divisor.
• Multiply the entire divisor by this quotient and subtract from the original dividend.
• Continue the process with the remaining polynomials.

Synthetic division, as described earlier, is a more streamlined method for dividing by binomials of the form \((x-a)\). Both methods provide insights not just into factors but also into depressions of degrees and potential roots for the polynomial in question.

Binomial Theorem

The Binomial Theorem is a formula for expanding powers of binomials. It shows us how expressions of the form \((x+a)^n\) can be expanded to a polynomial with coefficients coming from Pascal's Triangle or combinations.For example:

• \((x + a)^2 = x^2 + 2ax + a^2\)
• \((x + a)^3 = x^3 + 3ax^2 + 3a^2x + a^3\)

This theorem is fundamental when dealing with powers of binomials in polynomials, offering a systematic way to expand such expressions. While it seems detached from polynomial division, understanding each term's expansion aids in grasping the nature of polynomial multiplication and division.

Quadratic Factoring

Quadratic factoring is an essential skill in algebra, used to break down quadratic expressions into products of binomials. When you factor a quadratic, you aim to express it as \((x - p)(x - q)\), where \(p\) and \(q\) are the roots of the quadratic equation.For instance, if you have a quadratic expression like \(x^2 - 6x + 8\), you want two numbers that multiply to give 8 (the constant term) and add to give -6 (the linear coefficient). These numbers are -2 and -4, making the factors \((x-2)(x-4)\).Quadratic factoring is crucial when you have already divided your polynomial and need to factor the quotient further, such as in the final steps of completely factoring a polynomial. Being able to factor efficiently helps streamline the process and ensures completeness.