Question

divide
$x \: cube \: + 3x { }^{2} + 3x + 1 \:by \: x + \pi$
​

Answer 1

Answer:

The remainder is: $(1 - \pi)^3$, and the quotient is: $x^2 + (3 - \pi)x + \pi^2 - 3\pi + 3$

Step-by-step explanation:

The given expression is simply the expansion of:

$(x + 1)^3$

According to the remainder theorem, substituting x with the value obtained on equating $x + \pi$ with zero, shall give the remainder for the given polynomial.

So, putting $x = -\pi$ in the simplified equation, we get:

$(-\pi + 1)^3$ or $(1 - \pi)^3$ as the remainder.

Now, we know, that the given polynomial can be expressed as:

P(x) = Q(x)D(x) + R(x)

Where P(x) is the given polynomial, Q(x) is the quotient, D(x) the divisor, and R(x) the remainder. (if more elaboration is needed here, please comment)

Dividing the equation by D(x), we get:

$\frac{P(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}$

Or, $Q(x) = \frac{P(x) - R(x)}{D(x)}$

The above is a simple to understand statement: the quotient on the division of a quantity by a divisor, is the exact number obtained when the remainder itself is subtracted from the initial number, and then the division is done. This is natural, since the remainder is eliminated from the number when it is subtracted, so the number shall now be perfectly divisible by the divisor.

We currently know:

$P(x) = (x + 1)^3$

$D(x) = x + \pi$

$R(x) = (1 - \pi)^3$

So, for Q(x):

$Q(x) = \frac{(x + 1)^3 - (1 - \pi)^3}{x + \pi}$

Using $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$, it is factorized as:

$Q(x) = \frac{(x + 1 - (1 - \pi))((x + 1)^2 + (x + 1)(1 - \pi) + (1 - \pi)^2)}{x + \pi} = (x + 1)^2 + (x + 1)(1 - \pi) + (1 - \pi)^2$

$= x^2 + 1 + 2x + x - \pi x + 1 - \pi + \pi^2 + 1 - 2\pi$

$= x^2 + (3 - \pi)x + \pi^2 - 3\pi + 3$

So: $Q(x) = x^2 + (3 - \pi)x + \pi^2 - 3\pi + 3$