Advertisements
Advertisements
प्रश्न
If (x – 2) is a factor of the expression 2x3 + ax2 + bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b.
If (x – 2) is a factor of 2x3 + ax2 + bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b.
Advertisements
उत्तर
Since (x – 2) is a factor of polynomial 2x3 + ax2 + bx – 14, we have
2(2)3 + a(2)2 + b(2) – 14 = 0
⇒ 16 + 4a + 2b – 14 = 0
⇒ 4a + 2b + 2 = 0
Dividing the entire equation by 2,
⇒ 2a + b = –1 ...(1)
On dividing by (x – 3), the polynomial 2x3 + ax2 + bx – 14 leaves remainder 52,
2(3)3 + a(3)2 + b(3) – 14 = 52
⇒ 54 + 9a + 3b – 14 = 52
⇒ 9a + 3b = 52 – 40
⇒ 9a + 3b = 12
Dividing the entire equation by 3,
⇒ 3a + b = 4 ...(2)
Subtracting (1) and (2), we get
2a + b = –1
3a + b = 4
– – –
–a = –5
Substituting a = 5 in (1), we get
2 × 5 + b = –1
⇒ 10 + b = –1
⇒ b = –11
Hence, a = 5 and b = –11.
APPEARS IN
संबंधित प्रश्न
When x3 + 2x2 – kx + 4 is divided by x – 2, the remainder is k. Find the value of constant k.
Find ‘a‘ if the two polynomials ax3 + 3x2 – 9 and 2x3 + 4x + a, leave the same remainder when divided by x + 3.
Using the Remainder Theorem, factorise the following completely:
x3 + x2 – 4x – 4
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’.
Polynomials bx2 + x + 5 and bx3 − 2x + 5 are divided by polynomial x - 3 and the remainders are m and n respectively. If m − n = 0 then find the value of b.
Find the values of m and n when the polynomial f(x)= x3 - 2x2 + m x +n has a factor (x+2) and leaves a remainder 9 when divided by (x+1).
Find the value of p if the division of px3 + 9x2 + 4x - 10 by (x + 3) leaves the remainder 5.
Using remainder theorem, find the remainder on dividing f(x) by (x + 3) where f(x) = 2x2 – 5x + 1
When x3 – 3x2 + 5x – 7 is divided by x – 2,then the remainder is
If on dividing 2x3 + 6x2 – (2k – 7)x + 5 by x + 3, the remainder is k – 1 then the value of k is
