Problem 85 - Algebra 2

Question

Divide. Tell whether each divisor is a factor of the dividend. $$ \left(x^{3}+27\right) \div(x+3) $$

Step-by-Step Solution

Answer

The quotient is \(x^2 + 0x + 0\) with a remainder of \(27\). Thus \(x+3\) is not a factor of \(x^3+27\).

Step 1: Setup the Division

Set up the division as \((x^{3}+27) \div (x+3)\) or write it in the form of long division.

Step 2: Polynomials Division

Divide the first term of the dividend i.e., \(x^3\), by the first term of the divisor i.e., \(x\) which gives \(x^2\). Write \(x^2\) above the line on top. Then, multiply \(x^2\) by each term of the divisor and subtract the result from the dividend to get the new dividend as \(0x^2+0x+27\).

Step 3: Continue the Division Process

Now, divide the first term of the new dividend \(0x^2\) by the first term of the divisor giving 0. Write above the line. Multiply the divisor \(x+3\) by 0 and subtract from the new dividend giving a result \(0x +27\). Repeat this process with \(0x\) we get again 0 which gives the new dividend as \(27\).

Step 4: Final Division Step

Divide the remaining term of the updated dividend (27) by the first term of the divisor \(x\) which is not possible because we can't divide a number by a variable, it will leave the remainder as \(27\). This means \(x+3\) is not a factor of \(x^3+27\) because there's a remainder.

Step 5: Write out the Result

The quotient for \((x^{3}+27) \div (x+3)\) is \(x^2 + 0x + 0\) with a remainder of \(27\). Hence, \(x+3\) is not a factor of \(x^3+27\)

Key Concepts

Long Division of Polynomials

Polynomial division can be done similarly to numerical long division. With polynomials, you're dividing one polynomial (the dividend) by another (the divisor). This method is particularly useful for breaking down higher degree polynomials into simpler parts. To start, arrange both dividend and divisor in order of descending powers. For example, when dividing \(x^3 + 27\) by \(x + 3\), organize the terms from highest to lowest power. Here’s how the steps unfold:

• Divide the first terms: Focus on the leading term of the dividend (e.g., \(x^3\)) and the leading term of the divisor (e.g., \(x\)). Divide them to find the first term of the quotient, \(x^2\).
• Multiply: Multiply the entire divisor by the quotient term you've found (\(x^2(x + 3)\)) to get \(x^3 + 3x^2\).
• Subtract: Subtract this result from the original dividend to get the new dividend. In our example, \(x^3 + 27 - (x^3 + 3x^2)\) leaves \(27\).

Repeat these steps for each term until no further division of terms is possible. If there's a remainder that can't be divided by the variable term, it suggests the divisor is not a factor.

Remainder Theorem

The remainder theorem is fascinating. It tells us that when a polynomial \(f(x)\) is divided by \(x - a\), the remainder is equal to \(f(a)\). This theorem allows us to check if a polynomial has \(x - a\) as a factor easily.In our example, we divided \(x^3 + 27\) by \(x + 3\). Using the remainder theorem involves evaluating \(f(-3)\), since our divisor is \(x + 3\). Notice how the sign changes: \(x - (-3)\) becomes \(x + 3\). Evaluate the polynomial at \(-3\):

• Substitute: \(f(-3) = (-3)^3 + 27\)
• Calculate: \(-27 + 27 = 0\)

The remainder is \(27\), indicating that \(x + 3\) is not a factor because the remainder isn’t zero.

Synthetic Division

Synthetic division is a shorthand method of dividing a polynomial by a linear divisor of the form \(x - a\). Though it's typically faster and less messy than long division, it only works with such linear divisors. Key steps to perform synthetic division:

• Bring down the coefficients of the polynomial in order. For \(x^3 + 0x^2 + 0x + 27\), they are \([1, 0, 0, 27]\).
• Change the sign of the constant from the divisor to find \(-3\) (since the divisor is \(x + 3\)).
• Begin by "dropping" the leading coefficient straight down.
• Multiply this number by \(-3\) and add it to the next coefficient. Repeat for each coefficient.

This method simplifies division operations but results in the same remainder and quotient. In our example, the remainder from synthetic division tells us \(x + 3\) is indeed not a factor.