Question

if p(x) is =ax2+bx+cand a+c=b then one of the zeros is ​

Answer 1

Step-by-step explanation:

a+b+c=0⟹c=−a−b

ax2+bx+c=0

ax2+bx−a−b

(ax2−a)+(bx−b)=0

a(x2−1)+b(x−1)=0

a(x−1)(x+1)+b(x−1)=0

(x−1)(a(x+1)+b)=0

(x−1)(ax+a+b)=0

(x−1)(ax−c)=0

x=1;x=ca

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