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What must be added to [tex][tex]$x^3-6x^2+11x-8$[/tex][/tex] to make it a polynomial having a factor [tex][tex]$(x-3)$[/tex][/tex]?

Sagot :

To find what must be added to the polynomial [tex]\(x^3 - 6x^2 + 11x - 8\)[/tex] so that [tex]\((x - 3)\)[/tex] becomes a factor of the polynomial, we need to determine the remainder when the given polynomial is divided by [tex]\((x - 3)\)[/tex].

According to the remainder theorem, if we evaluate the polynomial at [tex]\(x = 3\)[/tex], the result will be the remainder when the polynomial is divided by [tex]\((x - 3)\)[/tex].

Given the polynomial [tex]\(P(x) = x^3 - 6x^2 + 11x - 8\)[/tex], we compute [tex]\(P(3)\)[/tex]:

[tex]\[ P(3) = 3^3 - 6 \cdot 3^2 + 11 \cdot 3 - 8 \][/tex]

First, we calculate the individual terms:

[tex]\[ 3^3 = 27 \][/tex]

[tex]\[ 6 \cdot 3^2 = 6 \cdot 9 = 54 \][/tex]

[tex]\[ 11 \cdot 3 = 33 \][/tex]

Next, we combine these results:

[tex]\[ P(3) = 27 - 54 + 33 - 8 \][/tex]

Now we simplify:

[tex]\[ 27 - 54 = -27 \][/tex]

[tex]\[ -27 + 33 = 6 \][/tex]

[tex]\[ 6 - 8 = -2 \][/tex]

The remainder when [tex]\(x^3 - 6x^2 + 11x - 8\)[/tex] is divided by [tex]\((x - 3)\)[/tex] is [tex]\(-2\)[/tex].

To make [tex]\((x - 3)\)[/tex] a factor of the polynomial, we need the remainder to be [tex]\(0\)[/tex]. Therefore, we must add [tex]\(2\)[/tex] to the polynomial [tex]\(x^3 - 6x^2 + 11x - 8\)[/tex].

Hence, the term that must be added is [tex]\(2\)[/tex]. So the updated polynomial will be:

[tex]\[ x^3 - 6x^2 + 11x - 8 + 2 = x^3 - 6x^2 + 11x - 6 \][/tex]

Therefore, [tex]\(2\)[/tex] must be added to [tex]\(x^3 - 6x^2 + 11x - 8\)[/tex] to make [tex]\((x - 3)\)[/tex] a factor of the polynomial.
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