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

Answer :

To solve the problem of finding the quotient of the polynomial [tex]\((x^4 + 5x^3 - 3x - 15)\)[/tex] divided by [tex]\((x^3 - 3)\)[/tex], we can use polynomial long division. Here's how you can do it step by step:

1. Set up the division: Write down the dividend [tex]\(x^4 + 5x^3 - 3x - 15\)[/tex] and the divisor [tex]\(x^3 - 3\)[/tex].

2. Divide the leading terms: Divide the leading term of the dividend, [tex]\(x^4\)[/tex], by the leading term of the divisor, [tex]\(x^3\)[/tex]. This gives you the first term of the quotient, which is [tex]\(x\)[/tex].

3. Multiply and subtract: Multiply the entire divisor [tex]\(x^3 - 3\)[/tex] by the first term of the quotient ([tex]\(x\)[/tex]) and subtract the result from the original dividend. This process gets rid of the leading term in the original dividend:
- Multiply: [tex]\(x \cdot (x^3 - 3) = x^4 - 3x\)[/tex].
- Subtract from the original dividend: [tex]\((x^4 + 5x^3 - 3x - 15) - (x^4 - 3x) = 5x^3 + 0x^2 - 15\)[/tex].

4. Repeat the process: Now with the new polynomial [tex]\(5x^3 + 0x^2 - 15\)[/tex], divide the leading term [tex]\(5x^3\)[/tex] by [tex]\(x^3\)[/tex] to get the next term in the quotient, which is [tex]\(+5\)[/tex].

5. Multiply and subtract again: Multiply the entire divisor by this new quotient term ([tex]\(5\)[/tex]) and subtract:
- Multiply: [tex]\(5 \cdot (x^3 - 3) = 5x^3 - 15\)[/tex].
- Subtract from the new polynomial: [tex]\((5x^3 + 0x^2 - 15) - (5x^3 - 15) = 0\)[/tex].

6. End of division: Since the remainder is zero and there are no more terms to drop down, the division is complete. The quotient is [tex]\(x + 5\)[/tex].

The quotient of [tex]\((x^4 + 5x^3 - 3x - 15)\)[/tex] divided by [tex]\((x^3 - 3)\)[/tex] is therefore [tex]\(x + 5\)[/tex].

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