Question

If f, of, x, equals, x, to the power 5 , plus, 5, x, to the power 4 , plus, 4f(x)=x
5
+5x
4
+4, then what is the remainder when f, of, xf(x) is divided by x, plus, 3x+3?

Answer 1

To solve the problem, we first need to clarify the function \( f(x) \) and its relationship to the given expression \( 4f(x) = x^5 + 5x^4 + 4 \). We'll first find the function \( f(x) \).

Starting from the equation:
\[
4f(x) = x^5 + 5x^4 + 4
\]
we can divide both sides by 4 to express \( f(x) \):
\[
f(x) = \frac{1}{4}(x^5 + 5x^4 + 4)
\]

Now, we need to find the expression for \( f(x) \) explicitly:
\[
f(x) = \frac{1}{4} x^5 + \frac{5}{4} x^4 + 1
\]

Next, we need to compute \( f(x)f(x) \). This means we will compute \( f(x) \times f(x) \). Let's denote \( f(x) \) simply as:
\[
f(x) = \frac{1}{4} x^5 + \frac{5}{4} x^4 + 1
\]

Now, we need \( f(x)f(x) \):
\[
f(x)f(x) = \left(\frac{1}{4} x^5 + \frac{5}{4} x^4 + 1\right) \left(\frac{1}{4} x^5 + \frac{5}{4} x^4 + 1\right)
\]

Now, we will simplify this expression. We can expand it using distributive property (i.e., the FOIL method) to find out \( f(x)f(x) \):
\[
f(x)f(x) = \left(\frac{1}{4} x^5\right)^2 + 2\left(\frac{1}{4} x^5\right)\left(\frac{5}{4} x^4\right) + 2\left(\frac{1}{4} x^5\right)(1) + \left(\frac{5}{4} x^4\right)^2 + 2\left(\frac{5}{4} x^4\right)(1) + 1^2
\]
Calculating the terms:
1. \( \left(\frac{1}{4} x^5\right)^2 = \frac{1}{16} x^{10} \)
2. \( 2\left(\frac{1}{4} x^5\right)\left(\frac{5}{4} x^4\right) = \frac{10}{16} x^9 = \frac{5}{8} x^9 \)
3. \( 2\left(\frac{1}{4} x^5\right)(1) = \frac{2}{4} x^5 = \frac{1}{2} x^5 \)
4. \( \left(\frac{5}{4} x^4\right)^2 = \frac{25}{16} x^8 \)
5. \( 2\left(\frac{5}{4} x^4\right)(1) = \frac{10}{4}x^4 = \frac{5}{2} x^4 \)
6. \( 1^2 = 1 \)

Combining these gives:
\[
f(x)f(x) = \frac{1}{16} x^{10} + \frac{5}{8} x^9 + \frac{25}{16} x^8 + \frac{1}{2} x^5 + \frac{5}{2} x^4 + 1
\]

Next, we need to find the remainder when \( f(x)f(x) \) is divided by \( x + 3 \). We can use the Remainder Theorem, which states that the remainder of \( p(x) \) when divided by \( x-c \) is \( p(c) \).

Let’s evaluate \( f(x)f(x) \) at \( x = -3 \):
\[
f(-3) = \frac{1}{4}(-3)^5 + \frac{5}{4}(-3)^4 + 1
\]
Calculating \( (-3)^5 = -243 \) and \( (-3)^4 = 81 \):
\[
f(-3) = \frac{1}{4}(-243) + \frac{5}{4}(81) + 1 = -\frac{243}{4} + \frac{405}{4} + 1 = \frac{162}{4} + 1 = \frac{162}{4} + \frac{4}{4} = \frac{166}{4} = \frac{83}{2}
\]
Thus,
\[
f(-3)f(-3) = \left(\frac{83}{2}\right)^2 = \frac{6889}{4}
\]

Finally, the remainder when \( f(x)f(x) \) is divided by \( x + 3 \) is:
\[
\frac{6889}{4}
\]

Therefore, the remainder is:
\[
\boxed{\frac{6889}{4}}
\]