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Question

Given that the zeroes of the cubic polynomial $$x^3 -  6x^2 + 3x + 10$$ are of the form $$a, a + b, a + 2b$$ for some real numbers $$a $$ and $$ b$$, find the values of $$a$$ and $$b$$.


Solution

Given cubic polynomial is 
$$p(x)=x^{ 3 }-6x^{ 2 }+3x+10$$
The zeros of the polynomial $$p(x)$$ are of the form $$a$$, $$a+b$$ and $$a+2b$$ 
Then,
   $$a+a+b+a+2b=-\frac{-6}{1}$$
$$=>3a+3b=6$$
$$=>a+b=2$$     ----------------(i)
Also,  $$a(a+b)+(a+b)(a+2b)+a(a+2b)=\frac{3}{1}$$
$$=>a^2+ab+a^2+2ab+ab+2b^2+a^2+2ab=3$$
$$=>3a^2+2b^2+6ab=3$$  ----(ii)
and   $$a(a+b)(a+2b)=-\frac{-10}{1}$$
$$=>a^3+a^2b+2a^2b+2ab^2=10$$
From (i), $$b=2-a$$
Putting this value in (ii), we get
$$=>3a^2+2(2-a)^2+6a(2-a)=3$$
$$=>3a^2+2(4-4a+a^2)+12a-a^2=3$$
$$=>-a^2+4a+5=0$$
$$=>a^2-4a-5=0$$
$$=>a^2-5a+a-5=0$$
$$=>a(a-5)+(a-5)=0$$
$$=>.(a-5)(a+1)=0$$
$$=>a=5$$  or   $$ a=-1$$
$$=>b=-3$$ or   $$ b=3$$ respectively
From equation (iii), we get
at $$a=5$$, $$b=-3$$
$$=>5^3+(5)^2(-3)+2(5)^2(-3)+2(5)(-3)^2=-10$$
$$=>125-75-150+90=-10$$
$$=>-10=-10$$ which is true.
and at $$a=-1$$, $$b=3$$, we have
$$=>(-1)^3+(-1)^2 3+2(-1)^2(3)+2(-1)(3)^2=-10$$
$$=>-1+3-6-18=10$$
$$=>-22=-10$$  which is not true.
Thus, $$a=5$$, $$b=-3$$
Zeros of the polynomial are $$5$$, $$5-3$$, $$5-2\times 3$$ ie $$5$$,$$2$$,$$-1$$

Mathematics

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