Question

x^2+21x-98 solve This please And get 100Points

Answer 1

♧♧HERE IS YOUR ANSWER♧♧

Now,

(x² + 21x - 98)

To find the zeroes of this expression, we consider it as an equation, which is

x² + 21x - 98 = 0

Now, to find the roots of this equation, please check out the attachment. That will help you.

*BY THE WAY*

May be, you want to know the roots of

x² + 21x + 98 = 0

=> x² + 14x + 7x + 98 = 0

=> x(x + 14) + 7(x + 14) = 0

=> (x + 14)(x + 7) = 0

So, either x + 14 = 0 or, x + 7 = 0

=> x = - 14 or, x = - 7

Therefore, the roots of the given equation are

x = - 14 and x = - 7.

♧♧HOPE THIS HELPS YOU♧♧

Answer 2

Answer:

$p(x) = {x}^{2} - 21x - 98$

for zero of polynomial,

let p(x)=0

${x}^{2} - 21x - 98 = 0 \\ {x}^{2} - 7x + 14x - 98 = 0 \\ x(x - 7) + 14(x - 7) = 0 \\ (x - 7)(x + 14) = 0 \\if \: x - 7 = 0 \\ x = 7 \\ if \: x + 14 = 0 \\ x = ( - 14)$

the zeroes of p(x) are 7 and (-14).

I hope it is helpful..........................