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
Using the Remainder Theorem, factorise each of the following completely.
3x3 + 2x2 – 23x – 30
The expression 4x3 – bx2 + x – c leaves remainders 0 and 30 when divided by x + 1 and 2x – 3 respectively. Calculate the values of b and c. Hence, factorise the expression completely.
Using Remainder Theorem, factorise : x3 + 10x2 – 37x + 26 completely.
If (x + 1) and (x – 2) are factors of x3 + (a + 1)x2 – (b – 2)x – 6, find the values of a and b. And then, factorise the given expression completely.
Using remainder Theorem, factorise:
2x3 + 7x2 − 8x – 28 Completely
In the following two polynomials. Find the value of ‘a’ if x + a is a factor of each of the two:
x3 + ax2 − 2x + a + 4
In the following two polynomials. Find the value of ‘a’ if x + a is a factor of each of the two:
x4 - a2x2 + 3x - a.
In the following two polynomials, find the value of ‘a’ if x – a is a factor of each of the two:
x6 - ax5 + x4 - ax3 + 3a + 2
Show that x2 - 9 is factor of x3 + 5x2 - 9x - 45.
If (x – 2) is a factor of 2x3 – x2 + px – 2, then
(i) find the value of p.
(ii) with this value of p, factorise the above expression completely
