College

The quotient of [tex]\left(x^4+5x^3-3x-15\right)[/tex] and [tex]\left(x^3-3\right)[/tex] is a polynomial. What is the quotient?

A. [tex]x^7+5x^6-6x^4-30x^3+9x+45[/tex]
B. [tex]x+5[/tex]
C. [tex]x-5[/tex]
D. [tex]x^7+5x^6+6x^4+30x^3+9x+45[/tex]

Answer :

To solve the problem, we need to find the quotient when dividing the polynomial [tex]\(x^4 + 5x^3 - 3x - 15\)[/tex] by the polynomial [tex]\(x^3 - 3\)[/tex]. This can be accomplished through polynomial division. Let's work through the steps:

1. Set up the division:

We are dividing [tex]\(x^4 + 5x^3 - 3x - 15\)[/tex] by [tex]\(x^3 - 3\)[/tex]. Write the dividend [tex]\(x^4 + 5x^3 - 3x - 15\)[/tex] under the division symbol and the divisor [tex]\(x^3 - 3\)[/tex] outside.

2. Divide the leading terms:

Look at the leading term of the dividend [tex]\(x^4\)[/tex] and the leading term of the divisor [tex]\(x^3\)[/tex]. Divide [tex]\(x^4\)[/tex] by [tex]\(x^3\)[/tex], which gives [tex]\(x\)[/tex].

3. Multiply and subtract:

Multiply the entire divisor [tex]\(x^3 - 3\)[/tex] by [tex]\(x\)[/tex], giving [tex]\(x^4 - 3x\)[/tex]. Subtract this from the original dividend:

[tex]\[
(x^4 + 5x^3 - 3x - 15) - (x^4 - 3x) = 5x^3 + 3x - 15
\][/tex]

4. Repeat the process:

Now, take the new polynomial [tex]\(5x^3 + 3x - 15\)[/tex] as the dividend. Divide the leading term [tex]\(5x^3\)[/tex] by the leading term [tex]\(x^3\)[/tex] of the divisor, which gives [tex]\(5\)[/tex].

5. Multiply and subtract again:

Multiply [tex]\(x^3 - 3\)[/tex] by [tex]\(5\)[/tex], giving [tex]\(5x^3 - 15\)[/tex]. Subtract this from [tex]\(5x^3 + 3x - 15\)[/tex]:

[tex]\[
(5x^3 + 3x - 15) - (5x^3 - 15) = 3x
\][/tex]

6. Finish up:

The remainder is [tex]\(3x\)[/tex], but since the remainder's degree is less than the divisor's degree, the division process stops here.

The polynomial that remains is [tex]\(x + 5\)[/tex]. Therefore, the quotient of [tex]\((x^4 + 5x^3 - 3x - 15)\)[/tex] divided by [tex]\((x^3 - 3)\)[/tex] is [tex]\(x + 5\)[/tex].

In conclusion, the answer is: [tex]\(x + 5\)[/tex].