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

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

1 answer

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 2x12, the quotient is x2+3x1 and the remainder is 88.

By the polynomial division theorem, we have:

f(x)2x12=(x2+3x1)+882x12

Multiplying both sides by the divisor 2x12:

f(x)=(x2+3x1)(2x12)+88

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

First, distribute (x2+3x1) by (2x12):

(x2+3x1)(2x12)=(x2)(2x)+(x2)(12)+(3x)(2x)+(3x)(12)+(1)(2x)+(1)(12)

Calculating each term:

  1. 2x3 from x22x
  2. 12x2 from x212
  3. 6x2 from 3x2x
  4. 36x from 3x12
  5. 2x from 12x
  6. 12 from 112

Now, combining like terms:

2x3+(12x2+6x2)+(36x2x)+12

This simplifies to:

2x36x238x+12

Next, we add the remainder 88:

f(x)=2x36x238x+12+88

This gives us:

f(x)=2x36x238x+100

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

f(x)=2x36x238x+100

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