A polynomial has the same remainder when divided by $x+k$ or $x-k$; what is $k$?

Question

Question

Given that $y = 3x^3 + 7x^2 - 48x + 49$ and that $y$ has the same remainder when it is divided by $x + k$ or $x - k$, find the possible values of $k$.

My attempt

Let $f(x) = 3x^3 + 7x^2 - 48x + 49$

$\text{Using Remainder Theorem,}$

\begin{align} f(-k) &= f(k) \\ 3(-k)^3 + 7(-k)^2 - 48(-k) + 49 &= 3(k)^3 + 7(k)^2 - 48(k) + 49 \\ -3k^3 - 7k^2 + 48k + 49 &= 3k^3 + 7k^2 - 48k + 49 \\ -3k^3 --3k^3 - 7k^2 - 7k^2 + 48k + 48k + 49 - 49 &= 0 \\ -6k^3 + 14k^2 + 96k &= 0 \\ \frac{-6k^3}{k} + \frac{14k^2}{k} + \frac{96k}{k} &= \frac{0}{k} \\ 6k^2 + 14k - 96 &= 0 \end{align}

$\text{Comparing } 6k^2 + 14k - 96 = 0 \text{ with } ak^2 + bk + c = 0, a = 6, b = 14, c = -96$

\begin{align} k = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a } &= \frac{ -(14) \pm \sqrt{(14)^2 - 4(6)(-96)} }{ 2(6) } \\ k &= 3 \text{ and } -5\frac{1}{3} \end{align}

$\therefore k = 3, -5\frac{1}{3} \text{ or } 0 $

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My answer is incorrect. The correct answer is $k = 0, 4 \text{ or } -4$

Answer 1

\begin{align} f(-k) &= f(k) \\ 3(-k)^3 + 7(-k)^2 - 48(-k) + 49 &= 3(k)^3 + 7(k)^2 - 48(k) + 49 \\ \bf-3k^3 - 7k^2 + 48k + 49 &= 3k^3 + 7k^2 - 48k + 49 \\ -3k^3 --3k^3 - 7k^2 - 7k^2 + 48k + 48k + 49 - 49 &= 0 \\ -6k^3 + 14k^2 + 96k &= 0 \\ \frac{-6k^3}{k} + \frac{14k^2}{k} + \frac{96k}{k} &= \frac{0}{k} \\ 6k^2 + 14k - 96 &= 0 \end{align}

Right there. $7(-k)^2 = 7k^2$. Looking through it, there are more sign errors throughout. $-7k^2 - 7k^2$ would be $-14k^2$, not $14k^2$.

Another one: $-3k^3 - -3k^3 = 0$.

Dividing by $k$ isn't exactly a kosher operation. What if $k = 0$?

Answer 2

Duncan Ramage has patiently showed you where you went wrong. I would like to provide some tips on how to avoid the errors you made and simplify your proof.

There is no need to write out every term of $f(k)$ and $f(-k)$. Instead you take the difference:

$$0=f(k)-f(-k)=3(k^3-(-k)^3) + 7(k^2-(-k)^2)-48(k-(-k))+(49-49)\\ =6k^3-96k=6k(k^2-16) \implies k=0, \pm 4.$$

When you get experienced you can skip the even-order terms altogether:

$$0=f(k)-f(-k)=3(k^3-(-k)^3) -48(k-(-k))\\ =6k^3-96k=6k(k^2-16) \implies k=0, \pm 4.$$

Answer 3

Your application of the Remainder Theorem does seem to be the approach for which the problem was designed. This is offered as a remark somewhat too long for a "comment", just to say a little about how your calculation can also be interpreted.

In writing the equation $ \ f(-k) \ = \ f(k) \ \ , $ we are in a sense asking for what values of $ \ x \ $ the function $ \ f(x) \ = \ 3x^3 + 7x^2 -48x + 49 \ \ $ [marked in blue in the graph above] "behaves" as an even-symmetry function. We can "resolve" our function into two "component" functions with even and odd symmetry, which is easily done for polynomial functions: $$ f(x) \ \ = \ \ f_{even}(x) \ + \ f_{odd}(x) $$ $$ \rightarrow \ \ 3x^3 + 7x^2 -48x + 49 \ \ = \ \ (7x^2 + 49) \ + \ (3x^3 - 48x) \ \ ; $$ $ f_{even}(x) \ \ \text{and} \ \ f_{odd}(x) \ $ are marked in green and red, respectively. For the even function, by definition, $ \ f(-k) \ = \ f(k) \ \ $ is true for all real numbers $ \ k \ \ , $ whereas this is only true for an odd function at its $ \ x-$ intercepts, where $ \ f(-k) \ = \ -f(k) \ = \ 0 \ \ . $

So we seek those values of $ \ x = k \ $ at which $ \ f_{odd}(k) = 0 \ \ ; $ thus, we can simply write

$$ 3k^3 - 48k \ = \ 0 \ \ \Rightarrow \ \ 3k · (k^2 - 16) \ = \ 0 \ \ \Rightarrow \ \ k \ = \ 0 \ , \ -4 \ , \ 4 \ \ . $$

The odd-symmetry function, then, can be thought of as a "filter" that only "passes" the values of $ \ f(x) \ $ that possess even symmetry. We see in the graph above that these are the marked points at which $ \ f(x) = f_{even}(x) \ $ and $ \ f_{odd}(x) = 0 \ \ . $ The solution $ \ k = 0 \ $ is "trivial" in that $ \ f(-0) = f(0) \ $ must be true for a function that is continuous at $ \ x = 0 \ \ . $

[A separate conclusion that we can make is that, except at $ \ x = 0 \ \ , $ there is no value $ \ x = k \ $ for which our function can have $ \ f(-k) = -f(k) \ \ , $ since $ \ f_{even} \ $ is never equal to zero. More generally, for a polynomial function, there is no value of $ \ x = k \ $ other than $ \ x = 0 \ $ at which $ \ f(-k) = -f(k) \ $ if the constant term of the polynomial is positive.]

Answer 4

By the factor theorem, $3x^3+7x^2-48x+49 = (ax+b)(x^2 - k^2) + c$. In other words, dividing by a quadratic term leaves a linear quotient and a constant remainder, for the powers on both sides to match.

Observe that there is only one way to make the terms of $x^3, x^2, x$. Thus $ax^3 = 3x^3 \implies x= 3$ and hence $-ak^2 = -48 \implies k^2 = 16, k = ±4$.