0.Given that the zeroes of the cubic polynomialx3–12x2+ 47x–60 are of the forma,a+b,a+ 2bfor somepositive real numbersaandb. Based on given information answer the following questions.
(i)The value ofais[1](1)3(2)4(3)5(4)6
(ii)The value ofbis[1](1)0(2)1(3)2(4)3
(iii)The sum of the zeroes of the polynomial is[1](1)10(2)–12(3)12(4)14
(iv)One of the zeroes of the polynomial is[1](1)–3(2)4(3)–4(4)6
Answers
Step-by-step explanation:
The zeroes of polynomial X^3 + 12 x^2 +47x-60 are in the form. a , a+ b and a+ 2b
Taking clue from q( iv) , by trial
P(3) = (3)^3 - 12(3)^2+47(3)-60
= 27- 12(9)+141-60
= 27-108+141-60
=168-168
= 0
so, x-3 is the factor of the above polynomial.
By dividing above polynomial by x-3 , we get
(x-3) (x^2-9x+20)
(x-3) (x^2-5x-4x+20) splitting middle term
(x-3)[ x ( x-5) - 4 (x-5)]
( x-3) (x-4) (x-5)
This gives us,
x-3 = 0 i.e x = 3
x-4= 0. x = 4
x-5 = 0. x= 5
The three zeroes are 3, 4 and 5
From this you can get,
a = 3 and b = 1
comparing with x= 3 ,4 and 5
a, a+ b. ,. a+2b
a= 3
a+ b = 3+1 = 4
a+ 2b = 3+ 2(1) = 3+2 = 5
I) value of. a is option (1)
ii) value of b is option (2)
iii) sum of zeroes is option (3)
i.e. 3+4+5 = 12
iv) one of the zero of the polynomial is option (2)