Question

Given that the zeroes of the cubic polynomial x3 – 12x + 47x – 60 are of the form a, a + b, a + 2b for some
positive real numbers a and b. Based on given information answer the following questions.
(i) The value of a is
(1) 3
(2) 4
(3) 5
(4) 6
(ii) The value of b is
(1) 0
(2) 1
(3) 2
(4) 3​

Answer 1

Answer:

the value of a is 5

the value of b is 6

Answer 2

The values of a and b are 3 and 1 respectively.

Polynomials are classified depending on the degree of its coefficient and is categorized as linear, quadratic, cubic etc. based on the highest degree of x.

The given equation is a cubic polynomial because the highest degree of x is 3. Also, when factorized and solved there will be 3 zeroes to this equation.

Similar to a quadratic equation in-order to get the solution to this cubic polynomial, that is finding the value of x, the equation needs to be factorized and then equated to zero.

Every cubic polynomial, P(x), is of the form:

$P(x)=Ax^{3} +Bx^{2} +Cx+D$ where $A\neq 0$

This can be factorized to be written in the form:$P(x)=A(x-\alpha )(x-\beta )(x-\gamma)$ where $\alpha, \beta \;and\;\gamma$ are constants.

Thus, the cubic polynomial has three zeroes to it represented by $\alpha, \beta \;and\;\gamma$. This means that the cubic polynomial has three roots to it which are the values of x

It has been derived that given the roots of this cubic equation to be a, b and c respectively then from the above general form of the cubic equation we have:

$a+b+c=-\frac{B}{A}------(1)$

$ab+bc+ca=\frac{C}{A}------(2)$

$abc=-\frac{D}{A}------(3)$

These are the three deduced expressions for a general cubic polynomial which represents the sum of the zeroes, product of the zeroes and a combination of both.

The cubic equation given here is $P(x)=x^{3} -12x^{2} +47x-60$ and its corresponding zeroes are given as  $a, a + b, a + 2b$.

From equations (1), (2) and (3) we can replace the given values to get:

$a=a\\b=a+b\\c=a+2b\\\\A=1\\B=-12\\C=47\\D=-60$

By substituting these values in the above derived expressions we have:

$a+(a+b)+(a+2b)=-\frac{(-12)}{1}$  [From (1)]

$a(a+b)+(a+b)(a+2b)+a(a+2b)=\frac{(47)}{1}$  [From (2)]

$a(a+b)(a+2b)=-\frac{(-60)}{1}-----(4)$  [From (3)]

By solving equation (1) we have:

$a+a+a+2b+b=12$

$\implies 3a+3b=12$

$\implies a+b=4-----(5)$

By substituting the above value from equation (5) in equation (4) we have:

$a(4)(a+2b)=60$

$\implies 4a(a+2b)=60$

$\implies a(a+2b)=\frac{60}{4}$

$\implies a(a+2b)=15$

On expanding we have:

$\implies a^{2} +2ab=15$

We know that, $(a+b)^{2}=a^{2}+b^{2}+2ab$

So to make it similar to the above equation we have:

$\implies a^{2} +2ab+b^{2}-b ^{2} =15$

$\implies (a+b)^{2}=15+b^{2}$

$\implies 4^{2}=15+b^{2}$

$\implies b^{2}=1$

$\implies b=1$

As b value is given as a  positive real number.

Now from the same equation (5) we have $b=4-a$. Hence by substituting the value of b in it we have:

$1=4-a$

$\implies a=4-1$

$\implies a=3$

So the values of a and b are 3 and 1 respectively.

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