Class 10 Exam > Class 10 Questions > For x2 + 2x + 5 to be a factor of x4 + &alpha...
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For x2 + 2x + 5 to be a factor of x4 + αx2 + β, the values of α and β should respectively be _______.
- a)2, 5
- b)5, 25
- c)6, 25
- d)5, 2
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
For x2 + 2x + 5 to be a factor of x4 + αx2 + β, the values ...
For x2 + 2x + 5 to be a factor of x4 + ax2 + b, remainder should be zero.
Now, remainder should be equal to zero.
∴ α – 6 = 0 and β – 25 = 0
⇒ α = 6 and β = 25
Now, remainder should be equal to zero.
∴ α – 6 = 0 and β – 25 = 0
⇒ α = 6 and β = 25
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Community Answer
For x2 + 2x + 5 to be a factor of x4 + αx2 + β, the values ...
To determine if x^2 + 2x + 5 is a factor of x^4 + ax^3 + bx^2 + cx + d, we need to perform polynomial division.
The division would look like this:
_____________________
x^2 + 2x + 5 | x^4 + ax^3 + bx^2 + cx + d
We start by dividing the first term of the dividend (x^4) by the first term of the divisor (x^2), which gives us x^2. We then multiply x^2 by the divisor (x^2 + 2x + 5), which gives us x^4 + 2x^3 + 5x^2.
Next, we subtract this product from the dividend:
x^4 + ax^3 + bx^2 + cx + d - (x^4 + 2x^3 + 5x^2)
This simplifies to:
ax^3 + (b - 2)x^2 + cx + d - 5x^2
We repeat the process by dividing the first term of the remaining expression (ax^3) by the first term of the divisor (x^2), which gives us a/x. We then multiply a/x by the divisor (x^2 + 2x + 5), which gives us a + 2ax + 5(a/x).
Next, we subtract this product from the remaining expression:
ax^3 + (b - 2)x^2 + cx + d - (a + 2ax + 5(a/x))
This simplifies to:
(b - 2 - 2a)x^2 + (c - 5(a/x))x + (d - 5a)
We repeat this process until we reach the end of the dividend.
For x^2 + 2x + 5 to be a factor, the remainder at the end of the division should be zero. In other words, we need the expression (b - 2 - 2a)x^2 + (c - 5(a/x))x + (d - 5a) to be equal to zero.
Therefore, for x^2 + 2x + 5 to be a factor of x^4 + ax^3 + bx^2 + cx + d, the coefficients (b - 2 - 2a), (c - 5(a/x)), and (d - 5a) must all equal zero.
The division would look like this:
_____________________
x^2 + 2x + 5 | x^4 + ax^3 + bx^2 + cx + d
We start by dividing the first term of the dividend (x^4) by the first term of the divisor (x^2), which gives us x^2. We then multiply x^2 by the divisor (x^2 + 2x + 5), which gives us x^4 + 2x^3 + 5x^2.
Next, we subtract this product from the dividend:
x^4 + ax^3 + bx^2 + cx + d - (x^4 + 2x^3 + 5x^2)
This simplifies to:
ax^3 + (b - 2)x^2 + cx + d - 5x^2
We repeat the process by dividing the first term of the remaining expression (ax^3) by the first term of the divisor (x^2), which gives us a/x. We then multiply a/x by the divisor (x^2 + 2x + 5), which gives us a + 2ax + 5(a/x).
Next, we subtract this product from the remaining expression:
ax^3 + (b - 2)x^2 + cx + d - (a + 2ax + 5(a/x))
This simplifies to:
(b - 2 - 2a)x^2 + (c - 5(a/x))x + (d - 5a)
We repeat this process until we reach the end of the dividend.
For x^2 + 2x + 5 to be a factor, the remainder at the end of the division should be zero. In other words, we need the expression (b - 2 - 2a)x^2 + (c - 5(a/x))x + (d - 5a) to be equal to zero.
Therefore, for x^2 + 2x + 5 to be a factor of x^4 + ax^3 + bx^2 + cx + d, the coefficients (b - 2 - 2a), (c - 5(a/x)), and (d - 5a) must all equal zero.
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