Question

fine the zeros of the polynomial p(x) =x^3-5x^2 -2x+24 if the product of its tow Zeros is 12​

Answer 1

$\bf{\Huge{\underline{\boxed{\bf{\red{ANSWER\::}}}}}}$

$\bf{\Large{\underline{\bf{Given\::}}}}$

The cubic polynomial p(x) = x³ - 5x² - 2x + 24. If the product of its two zeroes is 12.

$\bf{\Large{\underline{\bf{To\:find\::}}}}$

All the zeroes of the polynomial.

$\bf{\Large{\underline{\bf{\blue{Explanation\::}}}}}$

Let α , β and γ be the three zeroes.

We know that product of the zeroes in cubic polynomial:

$\longmapsto\sf{\alpha \beta \gamma\:=\:-\frac{Constant\:term}{Coefficient\:of\:x^{3} } =-\frac{d}{a} }$

According to the question:

$\leadsto\sf{12\gamma=-\frac{24}{1} }$

$\leadsto\sf{12\gamma=-24}$

$\leadsto\sf{\gamma=\cancel{\frac{-24}{12} }}$

$\leadsto\sf{\orange{\gamma\:=\:-2}}$

We know that sum of the zeroes in cubic polynomial:

$\longmapsto\sf{\alpha +\beta +\gamma\:=\:-\frac{Coefficient\:of\:x²}{Coefficient\:of\:x^{3} } =-\frac{b}{a} }$

$\leadsto\sf{\alpha +\beta +\gamma\:=\:-\frac{b}{a} }$

$\leadsto\sf{\alpha +\beta +(-2)\:=\:-\frac{-5}{1} }$

$\leadsto\sf{\alpha +\beta -2=5}$

$\leadsto\sf{\alpha +\beta =5+2}$

$\leadsto\sf{\alpha +\beta\: =\:7.......................(1)}$

We know that this formula:

$\mapsto\sf{\pink{(\alpha -\beta )^{2} =(\alpha +\beta )^{2} -4\alpha \beta }}$

Putting the value of equation(1) & product of two zeroes above that formula:

$\leadsto\sf{(\alpha -\beta )^{2} =\:(7)^{2} -4(12)}$

$\leadsto\sf{(\alpha -\beta )^{2} =\:49 \:-\:48}$

$\leadsto\sf{(\alpha -\beta )^{2} =1}$

$\leadsto\sf{\alpha -\beta \:=\:√1}$

$\leadsto\sf{\alpha -\beta \:=\:±1................................(2)}$

• $\bf{\large{\underline{\sf{For\:\alpha\: -\:\beta\: =\:1}}}}$

$\bf{\Large{\underline{\rm{Adding\:equation\:(1)\:&\:equation(2)\:,we\:get\::}}}}}$

$\mapsto\sf{\alpha \cancel{+\beta} +\alpha \cancel{-\beta} =7+1}$

$\mapsto\sf{2\alpha =8}$

$\mapsto\sf{\alpha =\cancel{\frac{8}{2} }}$

$\mapsto\sf{\orange{\alpha\: =\:4}}$

Putting the value of α in equation (1), we get;

$\mapsto\sf{4 +\beta =7}$

$\mapsto\sf{\beta =7-4}$

$\mapsto\sf{\orange{\beta\: =\:3}}$

__________________________________________________

• $\bf{\large{\underline{\sf{For\:\alpha\: -\:\beta\: =\:-1}}}}$

$\bf{\Large{\underline{\rm{Adding\:equation\:(1)\:&\:equation(2)\:,we\:get\::}}}}}$

$\mapsto\sf{\alpha \cancel{+\beta} +\alpha \cancel{-\beta} =7-1}$

$\mapsto\sf{2\alpha =6}$

$\mapsto\sf{\alpha =\cancel{\frac{6}{2} }}$

$\mapsto\sf{\orange{\alpha\: =\:3}}$

Putting the value of α in equation (1), we get;

$\mapsto\sf{3 +\beta =7}$

$\mapsto\sf{\beta =7-3}$

$\mapsto\sf{\orange{\beta\: =\:4}}$

Thus,

$\bf{\Large{\boxed{\sf{\purple{The\:three\:zeroes\:are\:3\:,\:4\:&\:-2}}}}}}}$

Answer 2

Solution:-

=> The polynomial

x³-5x²-2x+24

=> and compare with

ax³+bx²+cx+d

=> α+β+y= -b/a =5

=> αβy = -d/a = -24

12y = -24....(as given the product

of two zeros..

i.e. αβ=12

y= -2.

=> α+β+y=5

=> α+β-2=5

=> α+β =7. .................(1)

=> (α+β)²=7²

=> (α-β)²+4αβ=49

=> (α-β)²+4*12=49

=> (α-β)²+48= 49

=> (α-β)² =1

=> α-β = √1

=> α-β=1. ....................(2)

• subtracting (2) From (1)

=> α+β-(α-β)=7-1

=> α+β-α+β=6

=> 2β =6

=> β=3

• putting the value of β in the (2) eq

=> α-3=1

=> α=4

=> α=4,β=3,y= -2

i hope it helps you.