Polynomials Help

• Mar 30th 2009, 09:00 AM
ajmaal14
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.
• Mar 30th 2009, 10:04 AM
masters
Quote:

Originally Posted by ajmaal14
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,

\$\displaystyle f(1)=4\$

\$\displaystyle f(2)=-3\$

\$\displaystyle f(-2)=49\$

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

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

\$\displaystyle 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 \$\displaystyle ax^2+bx+c=0\$, we get:

\$\displaystyle 2x^2-13x+15=0\$
• Mar 30th 2009, 10:29 AM
HallsofIvy
[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 \$\displaystyle f(x)= ax^2+ bx+ c\$, divided by x-1, gives remainder 4, then \$\displaystyle f(1)= a+ b+ c= 4\$.
If \$\displaystyle f(x)= ax^2+ bx+ c\$, divided by x-2, gives remainder -3, then \$\displaystyle f(2)= 4a+ 2b+ c= -3\$.
If \$\displaystyle f(x)= ax^2+ bx+ c\$, divided by x+2= x-(-2), gives remainder 49, then \$\displaystyle 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 \$\displaystyle f(x)= 2x^2- 13x+ 15\$.

\$\displaystyle 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 \$\displaystyle 2x^2- 13x+ 15\$ divided by x+ 2 gives quotient 2x- 17 with remainder 49.

Quote:

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 \$\displaystyle f(x)= ax^2+ bx+ c\$, the two equations \$\displaystyle f(-2)= 4a- 2b+ c= -10\$ and \$\displaystyle 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 \$\displaystyle f(x)= ax^2+(3-a)x- (4+6a)\$ and divide that by \$\displaystyle (x+2)(x-3)= x^2- x- 6\$.