Is R(P+Q) for This Polynomial Problem 15?

  • Thread starter camilus
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In summary, the polynomial x^3+3x^2+9x+3 is a factor of x^4+4x^3+6Px^2+4Qx + R. By using long division of polynomials, the coefficients of each can be set equal to zero and resolved, or by letting the other factor be (x+a) and setting up equations relating the coefficients, the value of R(P+Q) can be determined. In this case, the value is 15.
  • #1
camilus
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if the polynomial [tex]x^3+3x^2+9x+3[/tex] is a factor of [tex]x^4+4x^3+6Px^2+4Qx + R[/tex], what is [tex]R(P+Q)[/tex]?
 
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  • #2
camilus, you need show your attempt before we can help.
 
  • #3
I already got the answer, I'm just trying to confirm it, see if I got the same answer other members get.
 
  • #4
So what's your answer?
 
  • #5
R(P+Q)=15

Have you tried the problem yet? What did you get?
 
  • #6
I don't get the same answer as you do. But if you can post your complete solution, it would be better, since that would definitely indicate if you (or I) made a numerical error somewhere, or made a mistake in an earlier step.
 
  • #7
Sorry, it was I who made a numerical error. I do get 15.
 
  • #8
Did you divide the first polynomial into the second?

then you could get the separate results and set them equal to zero, or you could multiply back and create equations.
 
  • #9
camilus said:
Did you divide the first polynomial into the second?
I didn't do that to solve the problem. But if you want to know if I double-checked the answer, then yes, it works out.

then you could get the separate results and set them equal to zero, or you could multiply back and create equations.

? I don't understand.
 
  • #10
Thats what I mean. Using long division of polynomials, the coefficients of each can be set equal to zero and resolved. Other than that, you can't multiply back the the (x+1) to the first polynomial and set the coefficients equal to the the coefficients of the second polynomial, and like earlier, just solve for P, Q, and R.
 
  • #11
camilus said:
Thats what I mean. Using long division of polynomials, the coefficients of each can be set equal to zero and resolved. Other than that, you can't multiply back the the (x+1) to the first polynomial and set the coefficients equal to the the coefficients of the second polynomial, and like earlier, just solve for P, Q, and R.

Hi camilus. My first instinct was to let the other factor be (x+a) and then just set up a few equations relating the coefficients.

That is,

r = 3a
4q = 9a+3
6p = 3a + 9
4 = a + 3


Since the last equation is trivial to solve (a=1) then solutions for r, p and q follow immediately.

So I got r = 3, q = 3 and p = 2, giving r(p+q) = 15
 

FAQ: Is R(P+Q) for This Polynomial Problem 15?

What is a simple polynomial problem?

A simple polynomial problem is an algebraic expression that involves only addition, subtraction, and multiplication of constants and variables raised to a non-negative integer power. These problems can be solved by applying the rules of algebra.

How do I solve a simple polynomial problem?

To solve a simple polynomial problem, you must first simplify the expression by combining like terms. Then, use the distributive property to remove parentheses. Finally, solve for the variable by isolating it on one side of the equation.

Can a simple polynomial problem have multiple solutions?

Yes, a simple polynomial problem can have multiple solutions. This is because there can be different values for the variable that satisfy the equation. These solutions can be found by using algebraic techniques such as factoring or the quadratic formula.

What is the degree of a simple polynomial problem?

The degree of a simple polynomial problem is the highest exponent of the variable in the expression. For example, in the expression 3x^2 + 5x + 2, the degree is 2. The degree is important because it determines the number of solutions a polynomial problem can have.

What are some real-life applications of simple polynomial problems?

Simple polynomial problems have many real-life applications, including calculating perimeter and area, determining profit and loss in business, and predicting the growth of populations or investments. They are also used in physics and engineering to model and solve problems involving motion and forces.

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