Advertisements
Advertisements
प्रश्न
x3 − 3x2 − 9x − 5
उत्तर
Let `f(x) = x^3 - 3x^2 - 9x -5 ` be the given polynomial.
Now, putting x = 1,we get
`f(-1) = (-1)^3 -3(-1)^ -9(-1) - 5`
`=-1 -3 +9 -5 = -9 +9 = 0`
Therefore, (x + 1)is a factor of polynomial f(x).
Now,
`f(x) = x^2 (x+1) -4x(x+1) -5(x +1)`
` = (x+1){x^2 -4x -5}`
` =(x+1){x^2 - 5x + x -5}`
` = (x+1)(x+1)( x-5)`
Hence (x+1) , (x+1) and (x - 5) are the factors of polynomial f(x) .
APPEARS IN
संबंधित प्रश्न
If `f(x)=2x^2-13x^2+17x+12` find `f(0)`
In each of the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the result by actual division: (1−8)
f(x) = x3 + 4x2 − 3x + 10, g(x) = x + 4
Show that (x + 4) , (x − 3) and (x − 7) are factors of x3 − 6x2 − 19x + 84
x4 − 7x3 + 9x2 + 7x − 10
2y3 + y2 − 2y − 1
2x4 − 7x3 − 13x2 + 63x − 45
If x − a is a factor of x3 −3x2a + 2a2x + b, then the value of b is
If (3x − 1)7 = a7x7 + a6x6 + a5x5 +...+ a1x + a0, then a7 + a5 + ...+a1 + a0 =
Factorise the following:
(p – q)2 – 6(p – q) – 16
(x + y)(x2 – xy + y2) is equal to
