Advertisements
Advertisements
प्रश्न
Using Remainder Theorem, factorise : x3 + 10x2 – 37x + 26 completely.
उत्तर
By remainder theorem,
For x = 1, the value of the given expression is the remainder.
x3 + 10x2 – 37x + 26
= (1)3 + 10(1)2 – 37(1) + 26
= 1 + 10 – 37 + 26
= 37 – 37
= 0
x – 1 is a factor of x3 + 10x2 – 37x + 26.
x2 + 11x – 26
`x - 1")"overline(x^3 + 10x^2 - 37x + 26)`
x3 – x2
11x2 – 37x
11x2 – 11x
– 26x + 26
– 26x + 26
0
∴ x3 + 10x2 – 37x + 26
= (x – 1)(x2 + 11x – 26)
= (x – 1)(x2 + 13x – 2x – 26)
= (x – 1)[x(x + 13) – 2(x + 13)]
= (x – 1)(x + 13)(x – 2)
∴ x3 + 10x2 – 37 + 26 = (x – 1)(x + 13)(x – 2)
APPEARS IN
संबंधित प्रश्न
Using the Remainder Theorem, factorise each of the following completely.
3x3 + 2x2 – 23x – 30
Given that x – 2 and x + 1 are factors of f(x) = x3 + 3x2 + ax + b; calculate the values of a and b. Hence, find all the factors of f(x).
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.
Using remainder Theorem, factorise:
2x3 + 7x2 − 8x – 28 Completely
Find the values of a and b in the polynomial f(x) = 2x3 + ax2 + bx + 10, if it is exactly divisible by (x+2) and (2x-1).
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:
x5 - a2x3 + 2x + a + 1.
Show that x2 - 9 is factor of x3 + 5x2 - 9x - 45.
If (x – a) is a factor of x3 – ax2 + x + 5; the value of a is ______.
For the polynomial x5 – x4 + x3 – 8x2 + 6x + 15, the maximum number of linear factors is ______.