QuestionProve that $2 x^{4}-6 x^{3}+3 x^{2}+3 x-2$ is divisible by $x^{2}-3 x+2$ without division.

Studdy Solution

STEP 1

Assumptions1. We have a polynomial $x^{4}-6 x^{3}+3 x^{}+3 x-$ . We are asked to prove that it is exactly divisible by $x^{}-3 x+$
3. We are not allowed to use actual division

STEP 2

First, we need to factorize the divisor polynomial $x^{2}- x+2$ .

STEP 3

The factorization of the polynomial $x^{2}-3 x+2$ can be done by finding two numbers that multiply to $2$ (the constant term) and add to $-3$ (the coefficient of $x$ ). These numbers are $-1$ and $-2$ .
So, we can write the divisor as $x^{2}-3 x+2 = (x-1)(x-2)$

STEP 4

Now, we need to prove that the given polynomial is divisible by both $(x-1)$ and $(x-2)$ .

STEP 5

Let's first check the divisibility by $(x-1)$ . We do this by substituting $x=1$ in the given polynomial. If the result is zero, then the polynomial is divisible by $(x-1)$ .

STEP 6

Substitute $x=1$ in the given polynomial $2(1)^{4}-6(1)^{3}+3(1)^{2}+3(1)-2 =2-6+3+3-2 =0$

STEP 7

Since the result is zero, the given polynomial is divisible by $(x-1)$ .

STEP 8

Next, we check the divisibility by $(x-2)$ . We do this by substituting $x=2$ in the given polynomial. If the result is zero, then the polynomial is divisible by $(x-2)$ .

STEP 9

Substitute $x=2$ in the given polynomial $2(2)^{4}-6(2)^{3}+3(2)^{2}+3(2)-2 =32-48+12+6-2 =$

STEP 10

Since the result is zero, the given polynomial is divisible by $(x-2)$ .

STEP 11

Since the given polynomial is divisible by both $(x-)$ and $(x-)$ , it is divisible by their product, which is the divisor $x^{}-3 x+$ .
Therefore, without actual division, we have proved that $x^{4}-6 x^{3}+3 x^{}+3 x-$ is exactly divisible by $x^{}-3 x+$ .

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