I'm working through identities but I can't figure out how to get further than multiplying out the above to get :
$$2x^3+3x^2-14x-5=2ax^3+3ax^2+3ax+bx^2+3bx+bx+3b+C$$
can someone give me a hint on what to do next?
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Good luck in finding such constants! Let $x=-1$. The right-hand side is $0$, and the left-hand side isn't.
Presumably there is a typo. Once it is fixed, we can go on.
Edit: The above was a response to the question before there was a $C$ on the right. Below is a solution of the corrected problem.
Put $x=-1$. Then the right-hand side is $C$, and the left-hand side is $10$, so $C=10$.
Put $x=0$. The right-hand side is $3b+10$, the left is $-5$. So $3b+10=-5$, and now we know $b$.
For $a$, the coefficient of $x^3$ on the left is $2$, and on the right it is $a$.
Now we need to verify that with the $a$, $b$, and $C$ we have found, the identity actually holds. This is necessary, for all we have shown is that if there are $a$, $b$, and $c$ that satisfy the equation, then $a$, $b$, $C$ must be as calculated. But perhaps there are no $a$, $b$, $C$ with the desired property.
Remark: Multiplying out and setting up a system of equations will get you there too, but it is somewhat tedious, and the probability of error increases.
André Nicolas
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yes there is sorry. – peter_gent Jun 12 '13 at 21:54
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why -1? i guess im bad at spotting these things but it seems arbitrary. – peter_gent Jun 13 '13 at 20:46
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The now deleted answer by amWhy used polynomial long division. Since $(x+3)(x+1)= x^2+4x+3$, if we perform the following polynomial long division
we get rightaway the quotient $2x-5=ax+b$, thus $a=2,b=-5$, and the remainder $10=C$.
have you any tips on spotting when to use this ?
Given two polynomials $A(x)$ and $B(x)$, with degree of $B(x)$ greater than $0$, we can find two other polynomials $Q(x)$ and $R(x)$ such that $$A(x)=B(x)Q(x)+R(x),$$ and the degree of $R(x)$ is lower than the degree of $B(x)$.
In your case $A(x)=2x^3+3x^2-14x-5$, $B(x)=x^2+4x+3$, $Q(x)=2x-5$ and $R(x)=10$.
Américo Tavares
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+1: Given the way the question was written this seems like the "intended" solution. – Steven Stadnicki Jun 12 '13 at 22:53
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@peter_gent This can be used when we have polynomials $A(x),B(x),Q(x),R(x)$ such that $A(x)=B(x)Q(x)+R(x)$. The integer division $a=bq+r$ is a particular case. – Américo Tavares Jun 12 '13 at 23:06
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By expanding and grouping like terms together, we obtain: $$ \begin{align*} 2x^3+3x^2-14x-5&=(ax+b)(x+3)(x+1) +C\\ 2x^3+3x^2-14x-5&=(ax+b)(x^2+4x+3) +C\\ 2x^3+3x^2-14x-5&=(ax)(x^2+4x+3)+b(x^2+4x+3) +C\\ 2x^3+3x^2-14x-5&=(ax^3+4ax^2+3ax)+(bx^2+4bx+3b) +C\\ 2x^3+3x^2-14x-5&=(a)x^3+(4a+b)x^2+(3a+4b)x+(3b+C) \end{align*} $$
Now let's compare coefficients. In particular, focus on the coefficients of $x^3$, $x^2$, and $x^0$ (the constant term). From this, we obtain: $$ \begin{align*} x^3 &: \boxed{2=a} \\ x^2 &: 3 = 4a+b \implies b = 3-4a=3-4(2) \implies \boxed{b=-5} \\ x^0 &: -5=3b+C \implies C=-5-3b=-5-3(-5) \implies \boxed{C=10} \end{align*} $$
Adriano
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So the coefficient of $x$ is $3a+4b=6-20/3=-2/3$ which is quite different from $-14$. – egreg Jun 12 '13 at 21:57
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@egreg Sorry, didn't see the edit. For $x^1$, we now have $3a+4b=3(2)+4(-5)=-14$, as desired. – Adriano Jun 12 '13 at 21:59
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I wanted only to warn you that the method was unsufficient; after the edit it works. – egreg Jun 12 '13 at 22:01
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Hint: look at the coefficients at $1$ and $x^3$.
TZakrevskiy
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@peter_gent I meant the term with no $x$ (or, as if you wish, $x^0$). $3b$ and $-5$ in your case. – TZakrevskiy Jun 12 '13 at 21:53
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yes, once you find $a$ and $b$ by this method, you have to check that other coefficients coincide, too. Apparently, you've made a typo in your polynomials, since $x=-1$ is a root for right side and is not for the left one. – TZakrevskiy Jun 12 '13 at 21:55