Question

plz solve this question immediately if u solve this question i will mark u as a brainlist​

Answer 1

$\large{\underline{\underline{\mathfrak{\green{\sf{Solution:-}}}}}}$.

$\large{\underline{\underline{\mathfrak{\green{\sf{Given\:here:-}}}}}}$.

• $\:x^3+x^2-17x+15\:=\:0$

• Zeros of this equation are 1,3,-5

$\large{\underline{\underline{\mathfrak{\bf{Explanation:-}}}}}$.

we know if x = a is a zeros of any equation ,

then , x =a exits this equations .

So, here , 1,3,-5 are zeros of this equation

So, X = 1,3,-5 exits this equations.

Case 1.

• When x = 1., keep in this equation

$\implies\:(1)^3+(1^2)+-17×(1)+15\:=\:0$

$\implies\:(1+1-17+15)\:=\:0$

$\implies\:(17-17)\:=\:0$

$\boxed{\implies\:0\:=\:0}$

_________________________

Case 2.

• when x=3, keep in equation

$\implies\:(3)^3+(3)^2-17×(3)+15\:=\:0$

$\implies\:(27+9-51+15)\:=\:0$

$\implies\:(51-51)\:=\:0$

$\boxed{\implies\:0\:=\:0}$

__________________,_____

Case 3.

• when x= -5 , Keep in equation

$\implies\:(-5)^3+(-5)^2-17×-5+15\:=\:0$

$\implies\:-125+25+85+15\:=\:0$

$\implies\:-125+125\:=\:0$

$\boxed{\implies\:0\:=\:0}$

_____________________

Here, x = 1,3,-5 are exits this equation ,

So,

• X = 1,3, -5 are zeros of this equation

$\boxed{\red{\:Thats\:Proved}}$

_____________________