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Answer :
We want to divide
[tex]$$
x^4 - 7x^3 + 5x - 45
$$[/tex]
by
[tex]$$
x-7.
$$[/tex]
Notice that the dividend is missing the [tex]$x^2$[/tex] term, so we rewrite it as
[tex]$$
x^4 - 7x^3 + 0x^2 + 5x - 45.
$$[/tex]
We will use polynomial long division.
─────────────────────────────
Step 1. Divide the leading term [tex]$x^4$[/tex] by [tex]$x$[/tex]:
[tex]$$
\frac{x^4}{x} = x^3.
$$[/tex]
This gives us the first term of the quotient. Now, multiply the divisor by [tex]$x^3$[/tex]:
[tex]$$
x^3 \cdot (x-7) = x^4 - 7x^3.
$$[/tex]
Subtract this from the dividend:
[tex]\[
\begin{array}{rcl}
& x^4 - 7x^3 + 0x^2 + 5x - 45\\[4mm]
- &(x^4 - 7x^3)\\ \hline
& 0x^4 + \,\,0x^3 + 0x^2 + 5x - 45.
\end{array}
\][/tex]
─────────────────────────────
Step 2. Bring down the next terms to work with [tex]$0x^2 + 5x - 45$[/tex]. The leading term here is [tex]$0x^2$[/tex]. Divide that by [tex]$x$[/tex]:
[tex]$$
\frac{0x^2}{x} = 0x.
$$[/tex]
Since it contributes nothing, we move on.
─────────────────────────────
Step 3. Now, divide the term [tex]$5x$[/tex] by [tex]$x$[/tex]:
[tex]$$
\frac{5x}{x} = 5.
$$[/tex]
This is the next term of the quotient. Multiply the divisor by 5:
[tex]$$
5 \cdot (x-7) = 5x - 35.
$$[/tex]
Subtract this from the current remainder:
[tex]\[
\begin{array}{rcl}
& 5x - 45\\[4mm]
- &(5x - 35)\\ \hline
& (5x - 5x) + (-45 + 35) = -10.
\end{array}
\][/tex]
─────────────────────────────
The remainder after the division is [tex]$-10$[/tex]. Thus, the quotient is
[tex]$$
x^3 + 5,
$$[/tex]
and the remainder is [tex]$-10$[/tex].
Finally, we can express the result as:
[tex]$$
x^4 - 7x^3 + 5x - 45 = (x-7)(x^3+5) - 10.
$$[/tex]
So the answer is:
- Quotient: [tex]$x^3+5$[/tex]
- Remainder: [tex]$-10$[/tex]
[tex]$$
x^4 - 7x^3 + 5x - 45
$$[/tex]
by
[tex]$$
x-7.
$$[/tex]
Notice that the dividend is missing the [tex]$x^2$[/tex] term, so we rewrite it as
[tex]$$
x^4 - 7x^3 + 0x^2 + 5x - 45.
$$[/tex]
We will use polynomial long division.
─────────────────────────────
Step 1. Divide the leading term [tex]$x^4$[/tex] by [tex]$x$[/tex]:
[tex]$$
\frac{x^4}{x} = x^3.
$$[/tex]
This gives us the first term of the quotient. Now, multiply the divisor by [tex]$x^3$[/tex]:
[tex]$$
x^3 \cdot (x-7) = x^4 - 7x^3.
$$[/tex]
Subtract this from the dividend:
[tex]\[
\begin{array}{rcl}
& x^4 - 7x^3 + 0x^2 + 5x - 45\\[4mm]
- &(x^4 - 7x^3)\\ \hline
& 0x^4 + \,\,0x^3 + 0x^2 + 5x - 45.
\end{array}
\][/tex]
─────────────────────────────
Step 2. Bring down the next terms to work with [tex]$0x^2 + 5x - 45$[/tex]. The leading term here is [tex]$0x^2$[/tex]. Divide that by [tex]$x$[/tex]:
[tex]$$
\frac{0x^2}{x} = 0x.
$$[/tex]
Since it contributes nothing, we move on.
─────────────────────────────
Step 3. Now, divide the term [tex]$5x$[/tex] by [tex]$x$[/tex]:
[tex]$$
\frac{5x}{x} = 5.
$$[/tex]
This is the next term of the quotient. Multiply the divisor by 5:
[tex]$$
5 \cdot (x-7) = 5x - 35.
$$[/tex]
Subtract this from the current remainder:
[tex]\[
\begin{array}{rcl}
& 5x - 45\\[4mm]
- &(5x - 35)\\ \hline
& (5x - 5x) + (-45 + 35) = -10.
\end{array}
\][/tex]
─────────────────────────────
The remainder after the division is [tex]$-10$[/tex]. Thus, the quotient is
[tex]$$
x^3 + 5,
$$[/tex]
and the remainder is [tex]$-10$[/tex].
Finally, we can express the result as:
[tex]$$
x^4 - 7x^3 + 5x - 45 = (x-7)(x^3+5) - 10.
$$[/tex]
So the answer is:
- Quotient: [tex]$x^3+5$[/tex]
- Remainder: [tex]$-10$[/tex]
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