Advertisements
Advertisements
प्रश्न
x3 − 23x2 + 142x − 120
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
Identify constant, linear, quadratic and cubic polynomials from the following polynomials
`p(x)=2x^2-x+4`
f(x) = 4x4 − 3x3 − 2x2 + x − 7, g(x) = x − 1
f(x) = x3 −6x2 − 19x + 84, g(x) = x − 7
Find α and β, if x + 1 and x + 2 are factors of x3 + 3x2 − 2αx + β.
x3 − 10x2 − 53x − 42
Find the remainder when x3 + 4x2 + 4x − 3 is divided by x.
If x + 1 is a factor of x3 + a, then write the value of a.
The value of k for which x − 1 is a factor of 4x3 + 3x2 − 4x + k, is
If both x − 2 and \[x - \frac{1}{2}\] are factors of px2 + 5x + r, then
Factorise the following:
(a + b)2 + 9(a + b) + 18
