Find the values of ‘a’ and ‘b’ so that the polynomial x3+ax2+bx-45 has (x-1)
and(x+5) as its factors.
Answers
Answer:
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Given :- Find the value of a and b so that the polynomial x³+ax²+bx-45 has x-1 and x+5 as its factors for the values of a and b as obtained above Factorise the given polynomial completely ?
Solution :-
when (x - 1) is factor of given polynomial . By remainder theorem we get,
→ f(x) = 0
→ f(1) = 0
→ (1)³ + a(1)² + b*1 - 45 = 0
→ 1 + a + b - 45 = 0
→ a + b - 44 = 0
→ a + b = 44 ----------- Eqn.(1)
similarly,
when (x + 5) is factor of given polynomial . By remainder theorem we get,
→ f(x) = 0
→ f(-5) = 0
→ (-5)³ + a(-5)² + b*(-5) - 45 = 0
→ (-125) + 25a - 5b - 45 = 0
→ 25a -5b - 170 = 0
→ 25a - 5b = 170
→ 5(5a - b) = 170
→ 5a - b = 34 ----------- Eqn.(2)
Adding Eqn.(1) and Eqn.(2) we get,
→ (5a - b) + (a + b) = 34 + 44
→ 5a + a - b + b = 78
→ 6a = 78
→ a = 13 .
putting value of a in Eqn.(1) ,
→ 13 + b = 44
→ b = 44 - 13
→ b = 31.
therefore, factorising the given polynomial by putting value of a and b , we get,
→ x³ + 13x² + 31x - 45
→ x³ - x² + 14x² - 14x + 45x - 45
→ x²(x - 1) + 14x(x - 1) + 45(x - 1)
→ (x - 1)(x² + 14x + 45)
→ (x - 1)(x² + 9x + 5x + 45)
→ (x - 1){x(x + 9) + 5(x + 9)}
→ (x - 1){(x + 9)(x + 5)}
→ (x - 1)(x + 5)(x + 9) (Ans.)
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