If p(x) is divided by (x-a)(x-b) prove that the remainder is
imprfctlyprfct:
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I think u forget to write p(x).........or it will be expand by identity
[(X-a)(x-b)=x^2+(-a-b)x+ab]
= x^2-ax-bx+ab and it will become p(x)
NOW ,
P(x) is divided by first g(x)=x-a
NOW,
It's remainder is 0
And if p(x) is divided by second g(x)=x-b
It's remainder is also 0.....
I think it will be correct
[(X-a)(x-b)=x^2+(-a-b)x+ab]
= x^2-ax-bx+ab and it will become p(x)
NOW ,
P(x) is divided by first g(x)=x-a
NOW,
It's remainder is 0
And if p(x) is divided by second g(x)=x-b
It's remainder is also 0.....
I think it will be correct
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