What is the sum of the coefficients?

Question

If $h(x) = x^4+ax^3+bx^2+cx+d$ then what is $a+b+c+d$?

I try: \begin{align} x=2: 2^4+2^3a+2^2b+2c+d = 3 &\implies 8a+4b+2c+d = -13 \label{I} \tag{I}\\ x=-2:-2^4-2^3a-2^2b-2c+d=3 &\implies -8a+4b-2c+d = -13 \label{II} \tag{II}\\ \eqref{I} + \eqref{II} \colon 8b +2d = -26 &\implies \boxed{4b+d = -13} \ em \tag{I}\\ 8a-13+2c=-13 &\implies 8a+2c = 0 \implies\boxed{4a+c = 0} \end{align}

I stop here..don't find another equation

If the graph of $;h;$ in your question is reliable then $;h;$ is an even function and thus it must be $;a=c=0;$ ... — DonAntonio, Mar 18 '21 at 12:18
Do you know derivatives and what they imply about function minima and maxima? — aschepler, Mar 18 '21 at 12:25
Or if not, $h(x)-3$ has two double roots, so we can determine its factorization up to a constant factor. — aschepler, Mar 18 '21 at 12:29

Arnaldo · Accepted Answer · 2021-03-18T12:45:16.660

3

Hint

I believe there is a better approach for this problem.

Considering the graph, the polinomial $h(x)-3$ has a double root at $-2$ and $2$. So, it can be writen as

$$h(x)-3=k(x-2)^2(x+2)^2=k(x^2-4)^2.$$

In addition, $h(1)$ give you the sum of the coefficients of $h(x)$.

Can you finish?

edited Mar 18 '21 at 12:45

answered Mar 18 '21 at 12:34

Arnaldo

21,342

Thank's but can you finish? I don't understand because h(1) give you the sum of the coefficients... – peta arantes Mar 18 '21 at 13:38
@petaarantes: Considering your polinomial, $h(1)=1+a+b+c+d$. Right? This is very usefull in many problems. If $p(x)$ is a polinomial, then $p(1)$ gives the sum of its coefficients. – Arnaldo Mar 18 '21 at 13:42
h(x) = k(x^4-8x^2+16) + 3... and the value of k?? – peta arantes Mar 18 '21 at 13:59
@petaarantes: only compare this polinomial you got with your original one. It gives you the value of $k$. More specifically, what is the value of the coefficient of $x^4$ on both polinomials? – Arnaldo Mar 18 '21 at 14:02
@petaarantes: Great! If you are satisfied, please select one answer (green mark). It avoids this question pop up many time. – Arnaldo Mar 18 '21 at 14:47

score 2 · Answer 2 · answered Mar 18 '21 at 12:25

By its graph $\;h\;$ is even and thus $\;a=c=0\;$. we're also given $\;h(2)=3\;$ and thus

$$3=h(2)=2^4+b\cdot2^3+d=16+8b+d\implies 8b+d=-13$$

...and I think some data is missing since it isn't possible to know from the given graph what is, for example, $\;h(0)=d\;$ ...unless you're allowed to use calculus and evaluate the first derivative, where it vanishes and etc. But you didn't tag this and didn't say anything about this...

Assuming you can use calculus: the first derivative vanishes at $\;x=2\;$ , thus

$$h'(2)=2\cdot2\cdot(2\cdot2^2+b)=0\implies b=-8$$

and then you the value of $\;d\;$ and we're done

What is the sum of the coefficients?

2 Answers2