What should be added to p(x) = x^{2} - 8x + 1 | vedclass

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(D) Let the polynomial be $p(x) = x^{2} - 8x + 10$. Let the value to be added be $k$. The new polynomial is $g(x) = x^{2} - 8x + 10 + k$. For $g(x)$ t

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$5$

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(D) Let the polynomial be $p(x) = x^{2} - 8x + 10$. Let the value to be added be $k$. The new polynomial is $g(x) = x^{2} - 8x + 10 + k$. For $g(x)$ to be divisible by $x - 3$,the remainder when $g(x)$ is divided by $x - 3$ must be $0$. According to the Remainder Theorem,the remainder is $g(3)$. Thus,$g(3) = 0$. Substituting $x = 3$ into $g(x)$: $(3)^{2} - 8(3) + 10 + k = 0$. $9 - 24 + 10 + k = 0$. $-15 + 10 + k = 0$. $-5 + k = 0$. Therefore,$k = 5$.

What should be added to $p(x) = x^{2} - 8x + 10$ so that the resulting polynomial is divisible by $x - 3$?

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