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   $$b=3$$ respectivelyFrom equation (iii), we getat $$a=5$$, $$b=-3$$$$=>5^3+(5)^2(-3)+2(5)^2(-3)+2(5)(-3)^2=-10$$$$=>125-75-150+90=-10$$$$=>-10=-10$$ 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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