Polynomial function help

12345678

Junior Member
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Mar 30, 2013
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102
Hello, I have just started polynomial functions and seem to be going wrong somewhere.
‘The polynomials f(x) and g(x) are 2X² + aX – 3 and 3X² - bX – 2 respectively, where a and b are constants. In the product f(x)g(x), the coefficient of X³ is 6, and the coefficient of X is 1. Find the coefficient of X².’
To tackle the problem, I first attempted to find a and b.
Finding b..
So I first multiplied g(x) by 2X² = 2X² ( 3X² - bX – 2) = 6X^4 – 2bX³ - 4X².
As the coefficient of X² is 6, 2b = 6, so b = 3.
g(x) = 3X² - 3X – 2.


Finding a..

I multiplied g(x) by aX = aX ( 3X² - 3X – 2 ) = 3aX³ - 3aX² - 2aX.

So 3a = 6, meaning a = 2

This is the first section that I am unsure on: a = 2 and b = 3, as I seem to get the wrong answer when finding the coefficient of X².
 

12345678

Junior Member
Joined
Mar 30, 2013
Messages
102
Sorry, it wouldn't let me post the whole question:
I will now do f(x)g(x) = (2X² + 2X – 3) ( 3X² - 3X – 2)

*2X² = +6X^4-6X³-4X²
*2X =+6X³-6X²-4X
*-3 =-9X²-9X+6
Total=6X^4-19X²-13X+6

So f(x)g(x) = 6X^4 – 19X² - 13X + 6

However, the coefficient of X² in my answer is 19, when the correct answer is 25.

Furthermore, if the coefficient of X³ is 6, should I not have 6X³ in my final answer?

This problem is seen when the coefficient of X = 1, and my answer has -13.

Any help in where I have gone wrong will be appreciated, cheers.
 

HallsofIvy

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Jan 27, 2012
Messages
4,990
Hello, I have just started polynomial functions and seem to be going wrong somewhere.
‘The polynomials f(x) and g(x) are 2X² + aX – 3 and 3X² - bX – 2 respectively, where a and b are constants. In the product f(x)g(x), the coefficient of X³ is 6, and the coefficient of X is 1. Find the coefficient of X².’
To tackle the problem, I first attempted to find a and b.
Finding b.
So I first multiplied g(x) by 2X² = 2X² ( 3X² - bX – 2) = 6X^4 – 2bX³ - 4X².
As the coefficient of X² is 6, 2b = 6, so b = 3.
You can't do just part of the multiplication! \(\displaystyle f(x)g(x)= (2x^2+ ax- 3)(3x^2- bx- 2)= 2x^2(3x^2- bx- 2)+ ax(3x^2- bx- 2)- 3(3x^2- bx- 2)= 6x^4- 2bx^3- 2x^2+ 3ax^3- abx^2- 2ax- 9x^2+ 3bx+ 6= 6x^4- (2b+3a)x^3- (2+ ab- 9)x^2- (2a- 3b)x+ 6= 0. NOW the coefficient of \(\displaystyle x^3\) is -2b- 3a= 6 and the coefficient of x is -2a+ 3b= 1. That gives you two equations to solve for a and b.\)\(\displaystyle
g(x) = 3X² - 3X – 2.
Finding a..

I multiplied g(x) by aX = aX ( 3X² - 3X – 2 ) = 3aX³ - 3aX² - 2aX.

So 3a = 6, meaning a = 2

This is the first section that I am unsure on: a = 2 and b = 3, as I seem to get the wrong answer when finding the coefficient of X².
\)
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
8,247
Sorry, it wouldn't let me post the whole question:
I will now do f(x)g(x) = (2X² + 2X – 3) ( 3X² - 3X – 2)

*2X² = +6X^4-6X³-4X²
*2X =+6X³-6X²-4X
*-3 =-9X²-9X+6
Total=6X^4-19X²-13X+6

So f(x)g(x) = 6X^4 – 19X² - 13X + 6

However, the coefficient of X² in my answer is 19, when the correct answer is 25.

Furthermore, if the coefficient of X³ is 6, should I not have 6X³ in my final answer?

This problem is seen when the coefficient of X = 1, and my answer has -13.

Any help in where I have gone wrong will be appreciated, cheers.
Have a look at this page.
 

12345678

Junior Member
Joined
Mar 30, 2013
Messages
102
You can't do just part of the multiplication! \(\displaystyle f(x)g(x)= (2x^2+ ax- 3)(3x^2- bx- 2)= 2x^2(3x^2- bx- 2)+ ax(3x^2- bx- 2)- 3(3x^2- bx- 2)= 6x^4- 2bx^3- 2x^2+ 3ax^3- abx^2- 2ax- 9x^2+ 3bx+ 6= 6x^4- (2b+3a)x^3- (2+ ab- 9)x^2- (2a- 3b)x+ 6= 0. NOW the coefficient of \(\displaystyle x^3\) is -2b- 3a= 6 and the coefficient of x is -2a+ 3b= 1. That gives you two equations to solve for a and b.[/COLOR][/SIZE][/FONT]
\)
\(\displaystyle
Thank you for your reply, but I cannot read you post after -(2a - 3b)x + 6.
Secondly, and I am probably wrong as you are more knowledgeable, but when you multiply f(x)g(x) I get: 6X^4 - 2bX^3 -4x^2 (you got 2x^2) + 3aX^3 - abx^2 - 2ax - 9x^2 + 3bx + 6
So when multiplying by 2x^2, you get 6X^4 - 2bX^3 - 4x^2 (2x^2 * -2) ; it is when multiplying g(x) by 2x^2 that we get different answers. If I am right I assume simplifying the answer will yield different results.
Also, do you use a programme or software to write in that easy to read bold font, it beats Microsoft word!
\)
 

12345678

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Joined
Mar 30, 2013
Messages
102
I’ve had another look at this question..
F(x)g(x) = (2X^2 + aX – 3) (3X^2 – bX – 2)
= 2X^2 (3X^2 – bX – 2) + aX (3X^2 – bX – 2) – 3(3X^2 – bX – 2)
= 6X^4 – 2bX^3 – 4X^2 + 3aX^3 – abX^2 – 2aX – 9X^2 + 3bX + 6
Now I have tried to simplify this.. : 6X^4 + X^3 (3a – 2b) + X^2 (-13 – ab ) + X (3b – 2a) + 6
However, I don’t know where to go from this – how do I find a and b?
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
8,247
I’ve had another look at this question..
F(x)g(x) = (2X^2 + aX – 3) (3X^2 – bX – 2)
= 2X^2 (3X^2 – bX – 2) + aX (3X^2 – bX – 2) – 3(3X^2 – bX – 2)
= 6X^4 – 2bX^3 – 4X^2 + 3aX^3 – abX^2 – 2aX – 9X^2 + 3bX + 6
Now I have tried to simplify this.. : 6X^4 + X^3 (3a – 2b) + X^2 (-13 – ab ) + X (3b – 2a) + 6
However, I don’t know where to go from this – how do I find a and b?
You know that
\(\displaystyle {\begin{array}{*{20}{r}}
{(3a - 2b) = 6}\\
{\,\left( { - 2a + 3b} \right) = 1}
\end{array}}\)

Solve that system.
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
3,434
Thank you for your reply, but I cannot read you post after -(2a - 3b)x + 6.
Secondly, and I am probably wrong as you are more knowledgeable, but when you multiply f(x)g(x) I get: 6X^4 - 2bX^3 -4x^2 (you got 2x^2) + 3aX^3 - abx^2 - 2ax - 9x^2 + 3bx + 6 The term in blue is incorrect. See below
So when multiplying by 2x^2, you get 6X^4 - 2bX^3 - 4x^2 (2x^2 * -2) ; it is when multiplying g(x) by 2x^2 that we get different answers. If I am right I assume simplifying the answer will yield different results.
Also, do you use a programme or software to write in that easy to read bold font, it beats Microsoft word!
Don't worry about the scripting language: it is fussy.

\(\displaystyle f(x) = 2x^2 + ax - 3\ and\ g(x) = 3x^2 - bx - 2 \implies f(x) * g(x) = (2x^2 + ax - 3)(3x^2 - bx - 2) =\)

\(\displaystyle f(x) = 2x^2(3x^2 - bx - 2) + ax(3x^2 - bx - 2) - 3(3x^2 - bx - 2) =\)

\(\displaystyle 6x^4 - 2bx^3 - 4x^2 + 3ax^3 - abx^2 - 2ax - 9x^2 + 3bx + 6 =\)

\(\displaystyle 6x^4 + (3a - 2b)x^3 - (13 + ab)x^2 + (3b - 2a)x + 6.\)

You did not expand the function correctly. Notice the coefficient of the x squared term.

By the terms of the problem

\(\displaystyle 3a - 2b = 6 \implies a = 2 + \dfrac{2b}{3}.\)

Also \(\displaystyle 3b - 2a = 1 \implies 3b - 2\left(2 + \dfrac{2b}{3}\right) = 1 \implies 3b - 4 - \dfrac{4b}{3} = 1 \implies\)

\(\displaystyle \dfrac{9b}{3} - \dfrac{4b}{3} = 4 + 1 = 5 \implies \dfrac{5b}{3} = 5 \implies b = \dfrac{3 * 5}{5} = 3 \implies\)

\(\displaystyle a = 2 + \dfrac{2 * 3}{3} = 4 \implies\)

\(\displaystyle 13 + ab = 13 + 4 * 3 = 25.\)

Now let's check.

\(\displaystyle (2x^2 + 4x - 3)(3x^2 - 3x - 2) = 2x^2(3x^2 - 3x - 2) + 4x(3x^2 - 3x - 2) - 3(3x^2 - 3x - 2) =\)

\(\displaystyle 6x^4 - 6x^3 - 4x^2 + 12x^3 - 12x^2 - 8x - 9x^2 + 9x + 6 = 6x^4 + 6x^3 - 25x^2 + x + 6.\)

You had the right idea. You made a careless mistake. To be honest, so did I the first time through. But checking my answer allowed me to find my mistake.
 

12345678

Junior Member
Joined
Mar 30, 2013
Messages
102
Thank you all for your reply's- I realised I was incorrectly solving the 2 equations to find a and b. Cheers again
 
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