When the function f, of, xf(x) is divided by 2, x, minus, 12x−1, the quotient is x, squared, plus, 3, x, minus, 1x

Question

2
+3x−1 and the remainder is 88. Find the function f, of, xf(x) and write the result in standard form.

Solve the problem just like the example below
⬇️
When the function
f, of, x
f(x) is divided by
x, minus, 2
x−2, the quotient is
x, squared, plus, x, plus, 1
x
2
+x+1 and the remainder is
minus, 5
−5. Find the function
f, of, x
f(x) and write the result in standard form.
start fraction, dividend, divided by, divisor, end fraction, equals, quotient, plus, start fraction, remainder, divided by, divisor, end fraction
divisor
dividend
​
=quotient+
divisor
remainder
​
start fraction, f, of, x, divided by, x, minus, 2, end fraction, equals, left parenthesis, x, squared, plus, x, plus, 1, right parenthesis, plus, start fraction, minus, 5, divided by, x, minus, 2, end fraction
x−2
f(x)
​
=(x
2
+x+1)+
x−2
−5
​
left parenthesis, x, minus, 2, right parenthesis, left parenthesis, start fraction, f, of, x, divided by, x, minus, 2, end fraction, right parenthesis, equals, left parenthesis, left parenthesis, x, squared, plus, x, plus, 1, right parenthesis, plus, start fraction, minus, 5, divided by, x, minus, 2, end fraction, right parenthesis, left parenthesis, x, minus, 2, right parenthesis
(x−2)(
x−2
f(x)
​
)=((x
2
+x+1)+
x−2
−5
​
)(x−2)
f, of, x, equals, left parenthesis, x, minus, 2, right parenthesis, left parenthesis, x, squared, plus, x, plus, 1, right parenthesis, minus, 5
f(x)=(x−2)(x
2
+x+1)−5

x, squared
x
2
plus, x
+x
plus, 1
+1
x
x
x, cubed
x
3
x, squared
x
2
x
x
minus, 2
−2
minus, 2, x, squared
−2x
2
minus, 2, x
−2x
minus, 2
−2
f, of, x, equals, left parenthesis, x, cubed, minus, x, squared, minus, x, minus, 2, right parenthesis, minus, 5
f(x)=(x
3
−x
2
−x−2)−5
f, of, x, equals, x, cubed, minus, x, squared, minus, x, minus, 7
f(x)=x
3
−x
2
−x−7

Answer 1

Let's solve the problem step by step, following the structure you've provided.

We are given that when the function \( f(x) \) is divided by \( 2x - 12 \), the quotient is \( x^2 + 3x - 1 \) and the remainder is \( 88 \).

By the polynomial division theorem, we have:

\[ \frac{f(x)}{2x - 12} = (x^2 + 3x - 1) + \frac{88}{2x - 12} \]

Multiplying both sides by the divisor \( 2x - 12 \):

\[ f(x) = (x^2 + 3x - 1)(2x - 12) + 88 \]

Now, we will expand the expression on the right side.

First, distribute \( (x^2 + 3x - 1) \) by \( (2x - 12) \):

\[ (x^2 + 3x - 1)(2x - 12) = (x^2)(2x) + (x^2)(-12) + (3x)(2x) + (3x)(-12) + (-1)(2x) + (-1)(-12) \]

Calculating each term:

1. \( 2x^3 \) from \( x^2 \cdot 2x \)
2. \( -12x^2 \) from \( x^2 \cdot -12 \)
3. \( 6x^2 \) from \( 3x \cdot 2x \)
4. \( -36x \) from \( 3x \cdot -12 \)
5. \( -2x \) from \( -1 \cdot 2x \)
6. \( 12 \) from \( -1 \cdot -12 \)

Now, combining like terms:

\[ 2x^3 + (-12x^2 + 6x^2) + (-36x - 2x) + 12 \]

This simplifies to:

\[ 2x^3 - 6x^2 - 38x + 12 \]

Next, we add the remainder \( 88 \):

\[ f(x) = 2x^3 - 6x^2 - 38x + 12 + 88 \]

This gives us:

\[ f(x) = 2x^3 - 6x^2 - 38x + 100 \]

In standard polynomial form, we write the final result as:

\[ \boxed{f(x) = 2x^3 - 6x^2 - 38x + 100} \]