Advertisements
Advertisements
प्रश्न
Factorise x3 + 6x2 + 11x + 6 completely using factor theorem.
Advertisements
उत्तर
Let f(x) = x3 + 6x2 + 11x + 6
For x = –1
f(–1) = (–1)3 + 6(–1)2 + 11(–1) + 6
= –1 + 6 – 11 + 6
= 12 – 12
= 0
Hence, (x + 1) is a factor of f(x).
x2 + 5x + 6
`x + 1")"overline(x^3 + 6x^2 + 11x + 6)`
x3 + x2
5x2 + 11x
5x2 + 5x
6x + 6
6x + 6
0
∴ x3 + 6x2 + 11x + 6 = (x + 1)(x2 + 5x + 6)
= (x + 1)(x2 + 2x + 3x + 6)
= (x + 1)[x(x + 2) + 3(x + 2)]
= (x + 1)(x + 2)(x + 3)
संबंधित प्रश्न
Factorise the expression f(x) = 2x3 – 7x2 – 3x + 18. Hence, find all possible values of x for which f(x) = 0.
Find the number that must be subtracted from the polynomial 3y3 + y2 – 22y + 15, so that the resulting polynomial is completely divisible by y + 3.
Show that (x – 1) is a factor of x3 – 7x2 + 14x – 8. Hence, completely factorise the given expression.
Using the factor theorem, show that (x - 2) is a factor of `x^3 + x^2 -4x -4 .`
Hence factorise the polynomial completely.
Find the value of a and b so that the polynomial x3 - ax2 - 13x + b has (x - 1) (x + 3) as factor.
If x – 2 is a factor of each of the following three polynomials. Find the value of ‘a’ in each case:
x5 - 3x4 - ax3 + 3ax2 + 2ax + 4.
Prove that (5x + 4) is a factor of 5x3 + 4x2 – 5x – 4. Hence factorize the given polynomial completely.
Use factor theorem to factorise the following polynomials completely: x3 – 19x – 30
If x3 – 2x2 + px + q has a factor (x + 2) and leaves a remainder 9, when divided by (x + 1), find the values of p and q. With these values of p and q, factorize the given polynomial completely.
Factorize completely using factor theorem:
2x3 – x2 – 13x – 6
