# Confused Polynomial

p(x) is a polynomial of degree 3 with p(1)=2, p(2)=3, p(3)=4, and p(4)=6. One of the factor of p(x+2) is ...

The answer is in the form of (x+a) with a is integer number Note by Dina Andini Sri Hardina
5 years, 10 months ago

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Hi Dina!

For problems like these consider a polynomial $H(x)=P(x)-(x+1)$, since $deg(P(x))=3$, so $deg(H(x))=3$. Plugging in $x=1,2,3$ in $H$, gives $0$ each time. Since nothing is known about the leading co-efficient of $P$, so let it be a real number $c$. So, $H(x)=c(x-1)(x-2)(x-3)$, giving $P(x)-(x+1)=c(x-1)(x-2)(x-3)$, now plug in $x=4$ in the above equation and using the fact that $P(4)=6$, $c=\frac{1}{6}$. So $P(x)=\frac{1}{6}(x-1)(x-2)(x-3)+(x+1)$. So $P(x+2)=\frac{x^{3}-x+6x+18}{6}$. You can now find the factor.................I hope this has helped you.

- 5 years, 10 months ago

very very help jit...anyway thanks:D

- 5 years, 10 months ago

but why do u use h(x)=p(x)-(x+1)?

- 5 years, 10 months ago

I took this polynomial because of the values given.

- 5 years, 10 months ago

We have 4 given values of the coefficients and 4 unknowns (the coefficients a, b, c and d). One could then use formal methods such as matrix inversion to get the value of the coefficients (a=1/6, b = -1, c=17/6 and d =0) Then p(x+2) = (x^2-5x+18)/6 which has a factor of (x+2).

Alternatively, one could just subtract the equation for p(1) from that for p(2) to remove d and get an equation in 3 variables.

p(1) = a + b + c + d = 2
p(2) = 8a + 4b + 2c + d = 3 p(3) = 27a + 9b + 3c + d = 4 p(4) = 64a + 16b + 4c + d = 6 (2)-(1) gives 7a + 3b + c = 1 (5) (3)-(2) gives 19a + 5b + c = 1 (6) (6) -(2) gives 12a + 2b = 0 2b = -12a b = -6a From (5), 7a -18a + c = 1 c = 11a+1 a + b + c + d = 2 a -6a + 11a + 1 + d = 2 6a + 1 + d = 2 d = 1-6a 64a + 16b + 4c + d = 6 64a -96a + 44a + 4 + 1-6a = 6 6a+5=6 6a=1 a = 1/6 b = -1 c = 11/6 + 1 1/6x^3 -x^2 + 17/6x x^3-6x^2+17x (x+2)^3 -6(x+2)^2 + 17(x+2) = x^3 + 8 + 6x(x+2) - 6(x^2+4x+4) + 17x + 34 = x^3+8+6x^2+12x - 6x^2-24x - 24 + 17x + 34 x^3 + 5x + 18

- 5 years, 10 months ago

i've tried this way but i already gave up

- 5 years, 10 months ago

Unfortunately, my solution not been presented / displayed as i would have liked and possibly, it looks confusing

- 5 years, 10 months ago

If $p(x)$ is cubic, we can generalize it as $p(x) = ax^3 + bx^2 + cx + d$. So, we can plug in our four data points to get four equations in four variables. Namely,

$a + b + c + d = 2$

$8a + 4b + 2c + d = 3$

$27a + 9b + 3c + d = 4$

$64a + 16b + 4c + d = 6$

We could solve these linear equations a whole host of ways. One of the easiest is to make this into a matrix and get it into reduced row eschelon form. Here's the Wolfram Alpha input for that: http://www.wolframalpha.com/input/?i=rref%28%5B1%2C+1%2C+1%2C+1%2C+2%5D%3B%5B8%2C+4%2C+2%2C+1%2C+3%5D%3B%5B27%2C+9%2C+3%2C+1%2C+4%5D%3B%5B64%2C+16%2C+4%2C+1%2C+6%5D%29

This gives us solutions of $a = \frac16, b = -1, c = \frac{17}{6}, d = 0$

Thus, $6p(x) = x^3 - 6x^2 + 17x$ -- note that we can multiply $p(x)$ by $6$ and still have the same roots.

This factors: $x(x^2 - 6x + 17)$. Thus, one factor is $x$, and that is indeed the only integer factor.

- 5 years, 10 months ago

That is the approach that i was trying to outline. But your presentation makes it easy to understand. But 6*p(x) = x^3+5x+18 with x+2 as one of the factors

- 5 years, 10 months ago

Sry...but what we have to calculate in the question?

- 5 years, 10 months ago

One of the factor of p(x+2) in the form x+a

- 5 years, 10 months ago

Jit has done it beautifully. I can also say that u should try to observe a pattern in the values given and then try to define a function.

- 5 years, 10 months ago

Thanks for the compliment mate, yeah observing the pattern is the key to such problems.

- 5 years, 10 months ago

One of the factors is x+2

- 5 years, 10 months ago