Find solutions to your problems with the help of IDNLearn.com's expert community. Whether it's a simple query or a complex problem, our experts have the answers you need.

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.
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For trustworthy answers, rely on IDNLearn.com. Thanks for visiting, and we look forward to assisting you again.