Class 10 Math All topics Polynomials

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=−1−6 $

$=>3a+3b=6$

Then,

$a+a+b+a+2b=−1−6 $

$=>3a+3b=6$

$=>a+b=2$ ----------------(i)

Also, $a(a+b)+(a+b)(a+2b)+a(a+2b)=13 $

$=>a_{2}+ab+a_{2}+2ab+ab+2b_{2}+a_{2}+2ab=3$

$=>3a_{2}+2b_{2}+6ab=3$ ----(ii)

$=>a_{2}+ab+a_{2}+2ab+ab+2b_{2}+a_{2}+2ab=3$

$=>3a_{2}+2b_{2}+6ab=3$ ----(ii)

and $a(a+b)(a+2b)=−1−10 $

$=>a_{3}+a_{2}b+2a_{2}b+2ab_{2}=10$

$=>a_{3}+a_{2}b+2a_{2}b+2ab_{2}=10$

From (i), $b=2−a$

Putting this value in (ii), we get

$=>3a_{2}+2(2−a)_{2}+6a(2−a)=3$

$=>3a_{2}+2(4−4a+a_{2})+12a−a_{2}=3$

$=>3a_{2}+2(2−a)_{2}+6a(2−a)=3$

$=>3a_{2}+2(4−4a+a_{2})+12a−a_{2}=3$

$=>−a_{2}+4a+5=0$

$=>a_{2}−4a−5=0$

$=>a_{2}−4a−5=0$

$=>a_{2}−5a+a−5=0$

$=>a(a−5)+(a−5)=0$

$=>a(a−5)+(a−5)=0$

$=>.(a−5)(a+1)=0$

$=>a=5$ or $a=−1$

$=>b=−3$ or $b=3$ respectively

$=>a=5$ or $a=−1$

$=>b=−3$ or $b=3$ respectively

From equation (iii), we get

at $a=5$, $b=−3$

$=>5_{3}+(5)_{2}(−3)+2(5)_{2}(−3)+2(5)(−3)_{2}=−10$

$=>125−75−150+90=−10$

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

and at $a=−1$, $b=3$, we have

$=>(−1)_{3}+(−1)_{2}3+2(−1)_{2}(3)+2(−1)(3)_{2}=−10$

$=>−1+3−6−18=10$

$=>−1+3−6−18=10$

$=>−22=−10$ which is not true.

Thus, $a=5$, $b=−3$

Zeros of the polynomial are $5$, $5−3$, $5−2×3$ ie $5$,$2$,$−1$

Zeros of the polynomial are $5$, $5−3$, $5−2×3$ ie $5$,$2$,$−1$

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