Advertisements
Advertisements
प्रश्न
If x3 + 6x2 + 4x + k is exactly divisible by x + 2, then k =
पर्याय
−6
−7
−8
−10
उत्तर
As f(x) = x3 + 6x2 + 4x + k is exactly divisible by (x +2).
i.e., ( x+2)is a factor of f(x).
Therefore, f(-2) =0
`(-2)^3 + 6(-2)^2 + 4(-2) + k = 0`
`-8 +24 - 8+ k = 0`
` = -8`
APPEARS IN
संबंधित प्रश्न
If the polynomials 2x3 + ax2 + 3x − 5 and x3 + x2 − 4x +a leave the same remainder when divided by x −2, find the value of a.
The polynomials ax3 + 3x2 − 3 and 2x3 − 5x + a when divided by (x − 4) leave the remainders R1 and R2 respectively. Find the value of the following case, if R1 = R2.
f(x) = 3x3 + x2 − 20x +12, g(x) = 3x − 2
x4 − 7x3 + 9x2 + 7x − 10
3x3 − x2 − 3x + 1
2y3 + y2 − 2y − 1
Write the remainder when the polynomialf(x) = x3 + x2 − 3x + 2 is divided by x + 1.
If x − a is a factor of x3 −3x2a + 2a2x + b, then the value of b is
One factor of x4 + x2 − 20 is x2 + 5. The other factor is
Factorise the following:
z² + 4z – 12
