Question

When P(x) is divided by (x - 1) and (x + 3), the remainders are 4 and 104 respectively. When P(x) is divided by x² - x + 1 the quotient is x² + x + 3 and the remainder is of the form ax + b. Find the remainder.

Answer 1

The remainder is 3x - 4

Explanation:

[Remember]
`(Dividend)/(Divisor) = Quotient + (Remainder)/(Divisor)`

So,
`Dividend = (Quotient)(Divisor) + Remainder`

In this case our dividend is always P(x).

Part 1

When the divisor is
`(x - 1)`, the remainder is
`4`, so we can say
`P(x) = (Quotient)(x - 1) + 4`

In order to get rid of "Quotient" from our equation, we must multiply it by 0, so
`(x - 1) = 0`

When solving for
`x`, we get

`x - 1 = 0\\x - 1 + 1 = 0 + 1\\x = 1`

When
`x = 1`,

`P(x) = (Quotient)(x - 1) + 4\\P(1) = (Quotient)(1 - 1) + 4\\P(1) = (Quotient)(0) + 4\\P(1) = 0 + 4\\P(1) = 4`

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Part 2

When the divisor is
`(x + 3)`, the remainder is
`104`, so we can say
`P(x) = (Quotient)(x + 3) + 104`

In order to get rid of "Quotient" from our equation, we must multiply it by 0, so
`(x + 3) = 0`

When solving for
`x`, we get

`x + 3 = 0\\x + 3 - 3 = 0 - 3\\x = -3`

When
`x = -3`,

`P(x) = (Quotient)(x + 3) + 104\\P(-3) = (Quotient)(-3 + 3) + 104\\P(-3) = (Quotient)(0) + 104\\P(-3) = 0 + 104\\P(-3) = 104`

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Part 3

When the divisor is
`(x^2 - x + 1)`, the quotient is
`(x^2 + x + 3)`, and the remainder is
`(ax + b)`, so we can say
`P(x) = (x^2 + x + 3)(x^2 - x + 1) + (ax + b)`

From Part 1, we know that
`P(1) = 4` , so we can substitute
`x = 1` and
`P(x) = 4` into
`P(x) = (x^2 + x + 3)(x^2 - x + 1) + (ax + b)`

When we do, we get:

`4 = (1^2 + 1 + 3)(1^2 - 1 + 1) + a(1) + b\\4 = (1 + 1 + 3)(1 - 1 + 1) + a + b\\4 = (5)(1) + a + b\\4 = 5 + a + b\\4 - 5 = 5 - 5 + a + b\\-1 = a + b\\a + b = -1`

We will call
`a + b = -1` equation 1

From Part 2, we know that
`P(-3) = 104`, so we can substitute
`x = -3` and
`P(x) = 104` into
`P(x) = (x^2 + x + 3)(x^2 - x + 1) + (ax + b)`

When we do, we get:

`104 = ((-3)^2 + (-3) + 3)((-3)^2 - (-3) + 1) + a(-3) + b\\104 = (9 - 3 + 3)(9 + 3 + 1) - 3a + b\\104 = (9)(13) - 3a + b\\104 = 117 - 3a + b\\104 - 117 = 117 - 117 - 3a + b\\-13 = -3a + b\\(-13)(-1) = (-3a + b)(-1)\\13 = 3a - b\\3a - b = 13`

We will call
`3a - b = 13` equation 2

Now we can create a system of equations using equation 1 and equation 2

`\left \{ {{a + b = -1} \atop {3a - b = 13}} \right.`

By adding both equations' right-hand sides together and both equations' left-hand sides together, we can eliminate
`b` and solve for
`a`

So equation 1 + equation 2:

`(a + b) + (3a - b) = -1 + 13\\a + b + 3a - b = -1 + 13\\a + 3a + b - b = -1 + 13\\4a = 12\\a = 3`

Now we can substitute
`a = 3` into either one of the equations, however, since equation 1 has less operations to deal with, we will use equation 1.

So substituting
`a = 3` into equation 1:

`3 + b = -1\\3 - 3 + b = -1 - 3\\b = -4`

Now that we have both of the values for
`a` and
`b`, we can substitute them into the expression for the remainder.

So substituting
`a = 3` and
`b = -4` into
`ax + b`:

`ax + b\\= (3)x + (-4)\\= 3x - 4`

Therefore, the remainder is
`3x - 4`.