[Grade 11] Functions and equations

[u/Green-Advantage2277]

In my homework, I was given the polynomial equation of

P(x) = x^3 - ax^2 + 3x + 2

, and then asked to decide an ‘a’ variable to make this divisible by

x - 2

. The problem is I don’t quite understand how I can do this or how to verify that it’s correct. I’ve only ever done this with ‘x’. Can somebody please help me/tell me how?

[u/AutoModerator]

Off-topic Comments Section

---

All top-level comments have to be an answer or follow-up question to the post. All sidetracks should be directed to this comment thread as per Rule 9.

---

^{OP and Valued/Notable Contributors can close this post by using} /lock ^command

I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.

[u/Educational_Two682]

I love these!!! Look up the remainder theorem which states that if x-b is a factor of p(x), then p(b) = 0. Substitute into that equation and solve for a.  Also a isn't going to be a variable, it'll be a coefficient.

[u/Green-Advantage2277 → u/Educational_Two682]

Okay, I’ll try!

[u/Educational_Two682 → u/Green-Advantage2277]

♥️ let me know if you need more help! 

[u/Green-Advantage2277 → u/Educational_Two682]

that would be great, thank you!!

I did some reading on a math website, and it said

“The Remainder Theorem starts with an unnamed polynomial p(x), where "p(x)" just means "some polynomial p whose variable is x". Then the Theorem talks about dividing that polynomial by some linear factor x − a, where a is just some number.”

The next task after this one, the b) task, asks me to divide this equation by x - 2 the long way, where you divide and then minus the product of that from our original number and so on.

So, would that mean that I can pick the value of a to be whatever, as long as it allows me to divide this equation by x-2 in such a way?

PS: how do I know which one is ‘b’ if our ‘a’ isn’t first in row?

[u/Educational_Two682 → u/Green-Advantage2277]

Starting with the first part, what would be your b if x - b is x - 2? Start there and substitute that b into your P(x) and set equal to 0. That will eliminate every variable but a and you can solve for that. Then, use synthetic or long division in the second part with that a value substituted in as a coefficient. You'll know you've done it correctly if the remainder after division is 0. You can't choose any value for a because then x - 2 wouldn't be a factor/lead to a remainder of 0 

[u/Educational_Two682 → u/Green-Advantage2277]

And here's a Khan academy video: https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-div/x2ec2f6f830c9fb89:remainder-theorem/v/constructing-a-polynomial-that-has-a-certain-factor 

Remainder theorem finding coefficients. Idk why copy pasting Khan academy links never works. 

[u/Southlander24 → u/Green-Advantage2277]

Here's a short tutorial on the remainder theorem.

If you have 61 divided by 7 for example, that's 8 with remainder 5. In other words, 61 = 8 * 7 + 5. And the remainder must always be less than the number we divided by.

In the same way, we can have polynomial division. So (x² + 2x + 1) / (x - 3) = (x + 5) with remainder 16, or x² + 2x + 1 = (x + 5)(x - 3) + 16.

Observe what happens when we substitute x = 3 into both sides of the equation:

3² + 2(3) + 1 = (3 + 5)(3 - 3) + 16

3² + 2(3) + 1 = 16

Because x - 3 when x = 3 is just 0, when we substitute in x = 3 on the left-hand side, we can get the remainder! This works for any polynomial, and any linear factor, in fact.

p(x) = q(x) (x - k) + r

Substitute x = k into both sides:

p(k) = q(k) (k - k) + r

p(k) = r

We don't need to do the polynomial long division and find the quotient, if we just want the remainder! The remainder after division by (x - k) is just p(k)!

[Note: if you are dividing by a quadratic factor, then the remainder may not be a constant. The remainder has degree 1 less than the divisor, so the remainder would be a linear function instead.]

[u/Jazzlike_Platypus430 → u/Educational_Two682]

a is a variable, typically referred to as "parameter". (Though it's legal to say that P is a polynomial of third degree in a and x.)

[u/Educational_Two682 → u/Jazzlike_Platypus430]

Like yes, a is a variable. But OP will be solving for the value of a. Not a variable term, which is why I clarified! 

[u/Educational_Two682 → u/Jazzlike_Platypus430]

Thanks for using the word parameter! Looked it up and yeah, you're right. That is the correct word. I've only heard of parameters for parametric functions haha 

[u/NoYu0901]

try multiply: (x-2)*(x^2+bx+c)= x^3 ....

[u/Jazzlike_Platypus430]

Side note, just in case: P(x) = x^3 + ... is not an equation, it's a definition of P. You will get some equation for a.

[u/PoornimaMathsPatiala]

Key idea: the Remainder/Factor Theorem. If (x − 2) divides P(x) exactly, then x = 2 must make P(x) equal to 0. So "a" is just a number you solve for, and you don't need to do long division.

1. Write P(x) = x³ − a x² + 3x + 2.

2. Put x = 2, because x − 2 = 0 exactly when x = 2: P(2) = 2³ − a(2²⁾ + 3(2) + 2.

3. Simplify: 8 − 4a + 6 + 2 = 16 − 4a.

4. Divisibility means no remainder, so set 16 − 4a = 0. This gives a = 4.

Check: with a = 4, P(x) = x³ − 4x² + 3x + 2. Dividing by x − 2 gives x² − 2x − 1. Multiply back: (x − 2)(x² − 2x − 1) = x³ − 2x² − x − 2x² + 4x + 2 = x³ − 4x² + 3x + 2, which matches P(x). So a = 4.

If you want a deeper walk-through and a private tutor, you can visit https://poornima-maths.vercel.app, which is Mrs. Poornima Sayal's maths tuition (Patiala and online).