If $2 x^{4} - 5 x^{3} + 2 x^{2} - x + a$ is exactly divisible by $x^{2} - 3 x + 2$ . Find $a$

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Solution

Let us first divide $2 x^{4} - 5 x^{3} + 2 x^{2} - x + a$ by $x^{2} - 3 x + 2$ as shown in the above image.

From the division, we observe that $2 x^{2} + x + 1$ is the quotient and $a - 2$ is the remainder.

Now, to find the value of $a$ , we equate the remainder to $0$ as shown below:

$a - 2 = 0 \Rightarrow a = 2$

Hence, $a = 2$ .

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Similar questions

2 x^{4} - 5 x^{3} + 2 x^{2} - x + 2

x^{2} - 3 x + 2

2 x^{4} - 5 x^{3} + 2 x^{2} - x + 2

x^{2} - 3 x + 2

2 x^{4} - 5 x^{3} + 2 x^{2} - x + 2

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f (x) = 2 x^{4} - 5 x^{3} + x^{2} + 3 x - 2

g (x)

q (x) = 2 x^{2} - 5 x + 3

r (x) = - 2 x + 1

g (x)

What must be added to $2 x^{4} - 5 x^{3} + 2 x^{2} - x - 3$ so that the result is exactly divisible by (x- 2)?

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