Advertisements
Advertisements
प्रश्न
x3 − 23x2 + 142x − 120
Advertisements
उत्तर
Let `f(x) = x^3 - 23x^2 + 142x -120` be the given polynomial.
Now, putting x=1,we get
`f(1) = (1)^3 - 23(1)^2 + 142(1) - 120`
` = 1 -23 + 142 - 120`
` = 143 - 143 = 0`
Therefore, (x-1)is a factor of polynomial f(x).
Now,
`f(x) = x^2(x-1) - 22x(x-1) + 120(x -1)`
`=(x-1){x^2 - 22x + 120}`
` = (x -1) {x^2 + 12x - 10x + 120}`
`=(x - 1)(x - 10)(x-12)`
Hence (x-1),(x-10) and (x-12) are the factors of polynomial f(x).
APPEARS IN
संबंधित प्रश्न
Write the coefficient of x2 in the following:
`pi/6x^2- 3x+4`
Identify polynomials in the following:
`g(x)=2x^3-3x^2+sqrtx-1`
Identify polynomials in the following:
`q(x)=2x^2-3x+4/x+2`
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
\[f(x) = 3 x^4 + 2 x^3 - \frac{x^2}{3} - \frac{x}{9} + \frac{2}{27}, g(x) = x + \frac{2}{3}\]
Find the remainder when x3 + 3x2 + 3x + 1 is divided by x + 1.
Find the remainder when x3 + 3x2 + 3x + 1 is divided by 5 + 2x .
Show that (x − 2), (x + 3) and (x − 4) are factors of x3 − 3x2 − 10x + 24.
If (x − 1) is a factor of polynomial f(x) but not of g(x) , then it must be a factor of
Factorise the following:
2a2 + 9a + 10
