If $p(x)=ax^3 -2x^2 +bx+c$, find $a, b$ and $c$ if $p(0)=12$, $p(-1)=3$ and $p(2)=36$

Question

Can someone please teach me how to do this question thanks!

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well, this question is from my text book. The concept I get from this is that you have to substitute the values into the polynomial. Although I have and I've only found c so far which is 12. — Red Queen10101, Jul 16 '13 at 07:26
This is basically the same question as the other one you posted. — m_l, Jul 16 '13 at 07:28
Doing all those substitutions turns this into a problem in linear algebra (a system of linear equations). Can you solve those? — Jyrki Lahtonen, Jul 16 '13 at 07:28
@m_l for some reason I tried the exact same thing as previous and it dosnt work... — Red Queen10101, Jul 16 '13 at 07:30
Show your work then, and someone will spot the error, I'm sure! — Daniel R, Jul 16 '13 at 07:32

score 3 · Accepted Answer · answered Jul 16 '13 at 07:49

3

Use $p(2)=36$ , you'll have $8a - 8 + 2b + 12 = 36$. Simplify it, you'll get $4a+b=16$. Together with the equation you've written down,

$a=3$,$b=4$

answered Jul 16 '13 at 07:49

crl0x7c2

106

Robert Lewis · Answer 2 · 2013-07-16T08:06:12.327

This is really a problem in basic linear algebra. If we simply plug in the three given values of the independent variable $x$, we obtain the three equations for $a$, $b$, $c$:

$p(0) = 12$ becomes

$c = 12$;

$p(-1) = 3$ becomes

$-a -b +c = 5$;

$p(2) = 36$ becomes

$8a + 2b + c = 44$;

looks like we get $c$ for free! Plugging $c= 12$ into the second and third equations yields

$-a -b = -7$;

$8a + 2b = 32$;

looks like we get

$6a = 18$,

$a = 3$;

and finally,

$b= 4$.

This method actually generalizes to arbitrary polynomials; the powers of the different values of $x$ form a matrix and the coefficients form a vector; the rest is clearly linear algabra. Glad to be of assistance in this matter. Cheers.

score 1 · Answer 3 · answered Jul 16 '13 at 07:51

So, you already found out that $c=12$. You also showed that $-a-b+10=3$, which we can simplify to $-a-b=-7$.

Doing a similar substitution, $$a(2)^3-2(2)^2+b(2)+c=36.$$

The arithmetic is not too difficult: $$8a-8+2b+c=36.$$

We can definitely move that $-8$ over, and remember how we already found out that $c=12$? Let's substitute that too: $$8a+2b+12=44.$$

Now for the 12: $$\color{blue}{8a+2b}=\color{blue}{32}.$$

This is good. Remember how we said that $-a-b=-7$? Let's just multiply that by $2$: $$\color{red}{-2a-2b}=\color{red}{-14}.$$

Now check this out: if we add these two previous equations together, the $b$-terms will cancel (i.e., $2b-2b=0$): $$\color{blue}{8a+2b}\color{red}{-2a-2b}=\color{blue}{32}\color{red}{-14}$$

or $$6a=18.$$

From here, it is very easy to find $a$, and now that you have two of the three variables, finding $b$ should be very easy after a bit of substitution. I hope this helps.

If $p(x)=ax^3 -2x^2 +bx+c$, find $a, b$ and $c$ if $p(0)=12$, $p(-1)=3$ and $p(2)=36$

3 Answers3