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Math Help - Polynomials Help

  1. #1
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    Polynomials Help

    I am stuck on 2 questions. I been trying for hours but its just not working.
    First question is: A quadratic equation f(x)=ax^2+bx+c, has the following properties when divided by x-1 the remainder is 4, when divided by x-2 the remainder is -3, and when divided by x+2 the remainder is 49

    for the question i did long division for each and i got 4a+2b+c=-3, c-2(b-2a)=49, and c+b+a=4 from there im stuck

    the second question is: when a polynomial is divided by x+2, the remainder is -10. When the same polynomial is divided by x-3, the remainder is 5. Determine the remainder when the polynomial is divided by (x+2)(x-3)

    Any help is greatly appreciated. Thanks guys.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by ajmaal14 View Post
    I am stuck on 2 questions. I been trying for hours but its just not working.
    First question is: A quadratic equation f(x)=ax^2+bx+c, has the following properties when divided by x-1 the remainder is 4, when divided by x-2 the remainder is -3, and when divided by x+2 the remainder is 49

    for the question i did long division for each and i got 4a+2b+c=-3, c-2(b-2a)=49, and c+b+a=4 from there im stuck

    the second question is: when a polynomial is divided by x+2, the remainder is -10. When the same polynomial is divided by x-3, the remainder is 5. Determine the remainder when the polynomial is divided by (x+2)(x-3)

    Any help is greatly appreciated. Thanks guys.
    Hi ajmaal14,

    I'll walk you through the 1st one. Then you can apply that model to the 2nd one.

    Using the remainder theorem,

    f(1)=4

    f(2)=-3

    f(-2)=49

    f(1)\Rightarrow a(1)^2+b(1)+c=4 \Rightarrow \boxed{{\color{red}a+b+c=4}}

    f(2)\Rightarrow a(2)^2+b(2)+c=-3 \Rightarrow \boxed{{\color{red}4a+2b+c=-3}}

    f(-2)\Rightarrow a(-2)^2+b(-2)+c=49 \Rightarrow \boxed{{\color{red}4a-2b+c=49}}

    Now, solve the three equation using your favorite method of solving a system of 3 equations in 3 unknowns. I used matrices and found a = 2, b = -13, and c = 15. You could use Cramer's Rule or substitution or something else.

    Substituting these back into ax^2+bx+c=0, we get:

    2x^2-13x+15=0
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  3. #3
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    [QUOTE=ajmaal14;291117]I am stuck on 2 questions. I been trying for hours but its just not working.
    First question is: A quadratic equation f(x)=ax^2+bx+c, has the following properties when divided by x-1 the remainder is 4, when divided by x-2 the remainder is -3, and when divided by x+2 the remainder is 49

    for the question i did long division for each and i got 4a+2b+c=-3, c-2(b-2a)=49, and c+b+a=4 from there im stuck [quote]
    Somewhat simpler than dividing is using this fact: if P(x) has remainder r when divided by x-a, then P(x)= (x-a)Q(x)+ r where Q is the quotient of the division. In particular P(a)= (a-a)Q(x)+ r= r.
    If f(x)= ax^2+ bx+ c, divided by x-1, gives remainder 4, then f(1)= a+ b+ c= 4.
    If f(x)= ax^2+ bx+ c, divided by x-2, gives remainder -3, then f(2)= 4a+ 2b+ c= -3.
    If f(x)= ax^2+ bx+ c, divided by x+2= x-(-2), gives remainder 49, then f(-2)= 4a- 2b+ c= 49.
    Yes, those are exactly the equations you give. Now you need to solve those equations for a, b, and c, by the usual method of solving systems of equations: removing one variable at a time. For example, if you subtract the second equation from the third, (4a- 2b+ c)- (4a+ 2b+ c)= 49+ 3, both b and c cancel and giving -4b= 52. Dividing both sides by -4, b= -13. If you subtract the first equation from the second, (4a+ 2b+ c)- (a+ b+ c)= -3- 4, the "c" terms cancel giving 3a+ b= -7. Since b= -13, 3a+ b= 3a- 13= -7. Adding 13 to both sides of the equation, 3a= 6 so a= 2. Finally put a= 2, b= -13, in any one of the equations, say a+ b+ c= 4, to get 2- 13+ c= 4, -11+ c= 4 or c= 15. a= 2, b= -13, c= 15 gives f(x)= 2x^2- 13x+ 15.

    2x^2- 13x+ 15 divided by x- 1 gives quotient x- 11 with remainder 4. [tex]2x^2- 13x+ 15]/math] divided by x- 2 gives quotient 2x- 9 with remainder -3, and 2x^2- 13x+ 15 divided by x+ 2 gives quotient 2x- 17 with remainder 49.

    the second question is: when a polynomial is divided by x+2, the remainder is -10. When the same polynomial is divided by x-3, the remainder is 5. Determine the remainder when the polynomial is divided by (x+2)(x-3)

    Any help is greatly appreciated. Thanks guys.
    Taking f(x)= ax^2+ bx+ c, the two equations f(-2)= 4a- 2b+ c= -10 and f(3)= 9a+ 3b+ c= 5 can be solved for two of the variables in terms of the third. For example, subtracting the first equation from the second, 5a+ 5b= 15 so a+ b= 3 and b= 3- a. Putting that into 4a- 2b+ c= -10, 4a- 2(3- a)+ c= 6a- 6+ c= -10 so c= -4- 6a.
    Because we do not have a third equation, we cannot determine specific values, but we can write f(x)= ax^2+(3-a)x- (4+6a) and divide that by (x+2)(x-3)= x^2- x- 6.
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