Advertisements
Advertisements
प्रश्न
If x – 2 is a factor of each of the following three polynomials. Find the value of ‘a’ in each case:
x2 - 3x + 5a
उत्तर
Let p(x) = x2 - 3x + 5a ...(i)
Since, (x - 2) is a factor of p(x), so p(2) = 0 ...(by factor theorem)
Put x = 2 in equation (i), we get
p(2) = (2)2 - 3 x 2 + 5a
= 4 - 6 + 5a
= 5a - 2
But p(2) = 0
⇒ 5a -2 = 0
⇒ 5a = 2
⇒ a = `(2)/(5)`.
APPEARS IN
संबंधित प्रश्न
A two digit positive number is such that the product of its digits is 6. If 9 is added to the number, the digits interchange their places. Find the number.
When divided by x – 3 the polynomials x3 – px2 + x + 6 and 2x3 – x2 – (p + 3) x – 6 leave the same remainder. Find the value of ‘p’.
Using the Remainder Theorem, factorise each of the following completely.
3x3 + 2x2 – 23x – 30
Factorise the expression f(x) = 2x3 – 7x2 – 3x + 18. Hence, find all possible values of x for which f(x) = 0.
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.
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.
The polynomial px3 + 4x2 – 3x + q is completely divisible by x2 – 1; find the values of p and q. Also, for these values of p and q, factorize the given polynomial completely.
Using the Reminder Theorem, factorise of the following completely.
2x3 + x2 – 13x + 6
Using the factor theorem, show that (x - 2) is a factor of `x^3 + x^2 -4x -4 .`
Hence factorise the polynomial completely.
If the polynomials ax3 + 4x2 + 3x - 4 and x3 - 4x + a leave the same remainder when divided by (x - 3), find the value of a.