1.

Check whether ‘6’ becomes a zero of x2 + 8x + 15 or not ? Give reasons.

Answer»

If p(k) = 0 then ‘k’ is a zero value of p(x) 

Now p(x) = x2 + 8x + 15 then 

p(6) = 62 + 8 (6) + 15 

= 36 + 48 + 15 ≠ 0 

As p(6) ≠ 0, ‘6’ cannot be the zero value of p(x).



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