Which of the following expressions is not a factor of [tex]$3x^3 - 10x^2 - 143x - 90$[/tex]?

Use long division to solve.

A. [tex]$x - 6$[/tex]
B. [tex][tex]$x - 9$[/tex][/tex]
C. [tex]$3x + 2$[/tex]
D. [tex]2 + 5$[/tex]



Answer :

Sure, let's use polynomial long division to determine which of the given expressions is not a factor of the polynomial [tex]\(3x^3 - 10x^2 - 143x - 90\)[/tex]. We'll divide the polynomial by each of the given expressions and check for a remainder. If the remainder is zero, then the expression is a factor; otherwise, it is not.

1. Division by [tex]\(x - 6\)[/tex]:

First, we set up the division of [tex]\(3x^3 - 10x^2 - 143x - 90\)[/tex] by [tex]\(x - 6\)[/tex].

[tex]\[ \begin{array}{r|rrr} & 3x^2 + 8x + 5 & \\ \hline x-6 & 3x^3 - 10x^2 - 143x - 90 \\ & -(3x^3 - 18x^2) \\ \hline & 8x^2 - 143x \\ & -(8x^2 - 48x) \\ \hline & -95x - 90 \\ & -(- 95x + 570) \\ \hline & -660 \\ \end{array} \][/tex]

The remainder is [tex]\(-660\)[/tex]. Since the remainder is not zero, [tex]\(x - 6\)[/tex] is not a factor.

2. Division by [tex]\(x - 9\)[/tex]:

Next, we set up the division of [tex]\(3x^3 - 10x^2 - 143x - 90\)[/tex] by [tex]\(x - 9\)[/tex].

[tex]\[ \begin{array}{r|rrr} & 3x^2 + 17x + 12 & \\ \hline x-9 & 3x^3 - 10x^2 - 143x - 90 \\ & -(3x^3 - 27x^2) \\ \hline & 17x^2 - 143x \\ & -(17x^2 - 153x) \\ \hline & -296x - 90 \\ & -(- 296x + 2670) \\ \hline & -90 \\ \end{array} \][/tex]

The remainder is 0. Since the remainder is zero, [tex]\(x - 9\)[/tex] is a factor.

3. Division by [tex]\(3x + 2\)[/tex]:

Next, we set up the division of [tex]\(3x^3 - 10x^2 - 143x - 90\)[/tex] by [tex]\(3x + 2\)[/tex].

[tex]\[ \begin{array}{r|rrr} & x^2 - 4x - 45 & \\ \hline 3x+2 & 3x^3 - 10x^2 - 143x - 90 \\ & -(3x^3 + 2x^2) \\ \hline & -12x^2 - 143x \\ & -(-12x^2 - 8x) \\ \hline & -151x - 90 \\ & -(- 151x - 22) \\ \hline & 0 \\ \end{array} \][/tex]

The remainder is 0. Since the remainder is zero, [tex]\(3x + 2\)[/tex] is a factor.

4. Division by [tex]\(x + 5\)[/tex]:

Finally, we set up the division of [tex]\(3x^3 - 10x^2 - 143x - 90\)[/tex] by [tex]\(x + 5\)[/tex].

[tex]\[ \begin{array}{r|rrr} & 3x^2 - 25x - 18 & \\ \hline x+5 & 3x^3 - 10x^2 - 143x - 90 \\ & -(3x^3 + 15x^2) \\ \hline & -25x^2 - 143x \\ & -(-25x^2 - 125x) \\ \hline & -18x - 90 \\ & -(- 18x - 90) \\ \hline & 0 \\ \end{array} \][/tex]

The remainder is 0. Since the remainder is zero, [tex]\(x + 5\)[/tex] is a factor.

Based on our calculations, [tex]\(x - 6\)[/tex] is not a factor of the polynomial [tex]\(3x^3 - 10x^2 - 143x - 90\)[/tex].