Given the equation $P(x) = x^4 + bx^2 + c$ and the points $(3,25)$ and $(1,9).$ How to find $b$ and $c$ such that it passes through the points?

Question

By using Desmos I've found that the values of $\;b = -8\;$ and $\;c = 16,\;$ but that doesn't really help me understand how to get those numbers.

Answer and why I got it wrong:

I've evaluated:

$ 9 = 1^4 + b\cdot1^2 + c \implies b = 8 - c \implies c = 8 - b$

and

$ 25 = 3^4 + b\cdot3^2 + c \implies c = -56 - 9b $

The part which I forgot to do was to substitute the equations into each other and not the original equation. KM101 provides a more complete breakdown.

$ c = -56 - 9(8 - c) \implies c = 16$

$ -9b - 56 = 8 - b \implies b = -8$

Answer 1

Substitute the $(x,y)$ pairs in to get a system of linear equations, which may be easily solved: $$1^4+1^2b+c=9\qquad3^4+3^2b+c=25$$ $$b+c=8\qquad 9b+c=-56$$

Answer 2

Hint:

$$25=3^4+b3^2+c \implies 9b+c=-56$$

$$9=1^4+b1^2+c \implies b+c=8$$

Answer 3

Here’s our function. $$f(x) = x^4+bx^2+c$$ We’re given two points. $$A: (1, 9)$$ $$B: (3, 25)$$ For point $A$, substitute $x$ and $y$ with $1$ and $9$, respectively. $$9 = 1^4+b(1)^2+c$$ $$9 = 1+b+c \implies \boxed{b+c = 8}$$ For point $B$, substitute $x$ and $y$ with $3$ and $25$, respectively. $$25 = 3^4+b(3)^2+c$$ $$25 = 81+9b+c \implies \boxed{9b+c = -56}$$ Now, use elimination to solve for $b$ and $c$. $$9b+c-(b+c)= -56-8$$ $$9b+c-b-c = -56-9 \implies 8b = -64$$ $$\boxed{b = -8}$$ Now, plug in $b = -8$ in either of the two equations to get $c$. $$9(-8)+c = -56$$ $$-72+c = -56 \implies \boxed{c = 16}$$ So, the final quartic equation containing these two points is $f(x) = x^4-8x^2+16$.

Answer 4

Notice that $P(u)-P(v)=u^4-v^4+b(u^2-v^2)$. From $25-9=81-1+b(9-1)$ you’ll find $b=-8$. Now exploit $P(1)=9$, that is $1-8+c=9$ to find $c=16$.