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what must be added to $x^3-6x^2+11x-8$ to make a polynomial having factor $x-3$?

If the required expression to be added be $K$ then $x^3-6x^2+11x-8+K$ is exactly divisible by $x-3$ but how do I find $K$??

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    Hint: being divisible by $x-3$ is equivalent to $3$ being a root of the polynomial. – 2017-01-28
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    Its remainder when dividing by $x-3$ is its value at 3, which is -2. So, adding 2 makes the trick. – 2017-01-28
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    $$-x^3 + 6x^2 - 11x + 8$$ – 2017-01-28
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    All you need to do is equate the function to 0 and substitute $x = 3$ – 2017-01-28
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    See https://en.wikipedia.org/wiki/Polynomial_remainder_theorem – 2017-01-28

5 Answers 5

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We need to add some constant $k $ such that the polynomial $p (x)+k $ is divisible by $(x-3) $, that is, having $3$ as one of its factors. Thus, $$p (3)+k =0$$ $$\Rightarrow (3)^3-6 (3)^2+11 (3) -8 +k =0$$ $$\Rightarrow \boxed {k = 2} $$ Hope it helps.

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Let $f(x)=x^3-6x^2+11x-8+K$

By factor theorem, $f(3)=0$

$27-54+33-8+K=0$

$K=2$

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Plug in $x=3$ into the polynomial you get $-2$, so you must add $2$ to it to make the polynomial zero at $3$.

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Let $f(x)=x^3-6x^2+11x-8$.

Hence, $f(x)=q(x)(x-3)+f(3)$, where $q$ is a polynomial with degree two.

Thus, we need to add $-f(3)=2$.

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I think you know when to add term and when to subtract term from polynomial.

So continue from that.

Remember that if we have given some factor then you can find value of variable from that factor and can put in polynomial. This is equal to either 0 if any remainder not given else equal to remainder.

We (x-3) as a factor. So x-3=0.

x=3.

Put value of x as 3 in polynomial and you have value of k.