Advertisements
Advertisements
Question
x4 + 10x3 + 35x2 + 50x + 24
Advertisements
Solution
Let \[f\left( x \right) = x^4 + 10 x^3 + 35 x^2 + 50x + 24\]
Now, putting x = 1,we get
`f(-1) = (-1)^4 + 10(-1)^3 + 35 (-1) +24`
` = 1 - 10 + 35 - 50 + 24 = 60 -60`
` = 0`
Therefore, (x +1)is a factor of polyno^2 + 50 (-1)mial f(x).
Now,
`f(x) = x^3 (x+1)9x^2(x +1) + 26x(x+1) + 24(x + 1)`
` = (x +1){x^3 +9x^2+ 26x + 24}`
`= (x +1)g(x) ..... (1)`
Where `g(x) = x^3 + 9x^2 + 26x +24`
Putting x = -2we get
`g(-2) = (-2)^3 + 9(-2)^2 +26 (-2)+ 24`
` = -8 + 36 - 52 + 24 = 60 -60`
` = 0`
Therefore, (x+2)is the factor of g(x).
Now,
`g(x) = x^2 (x+2) + 7x(x + 2) + 12(x + 2)`
` = (x + 2){x^2 + 7x + 12}`
`= (x +2)(x^2 + 4x + 3x + 12)`
` = (x + 2)(x+3)(x + 4) ........... (2)`
From equation (i) and (ii), we get
f(x) = (x + 1) (x + 2)(x+3)(x +4)
Hence (x + 1),(x + 2), (x + 3) and (x + 4 ) are the factors of polynomial f(x).
APPEARS IN
RELATED QUESTIONS
Write the degrees of the following polynomials
0
f(x) = x4 − 3x2 + 4, g(x) = x − 2
Show that (x − 2), (x + 3) and (x − 4) are factors of x3 − 3x2 − 10x + 24.
Find the values of p and q so that x4 + px3 + 2x3 − 3x + q is divisible by (x2 − 1).
y3 − 2y2 − 29y − 42
2y3 + y2 − 2y − 1
Factorise the following:
t² + 72 – 17t
Factorise the following:
(p – q)2 – 6(p – q) – 16
If (x + 5) and (x – 3) are the factors of ax2 + bx + c, then values of a, b and c are
Factorise:
x3 – 6x2 + 11x – 6
