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$$32x+32-8ax-4a-2a+2bx+b=0$$ Equating coefficients: $$32-8a+2b=0\tag1$$ $$32-6a+b=0\tag2$$ $(1)-2\times(2):$ $$-32+4a=0\\a=8\\b=16.$$

This is a bizarre method that my maths teacher uses to solve this linear algebra equation. Something about it seems wrong. Can anyone explain or confirm?

ryang
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Justin
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    If the equation given on the first line is a polynomial equality (i.e. holds "for all" $x$ where $x$ comes from the real numbers etc.) then what your teacher has done is basically use the fact that if a polynomial equals zero then all its coefficients equal zero. Then your teacher found the coefficient of $x$ by "collecting" the $x$ terms and set it to zero. They then found the constant term by taking the rest of the terms, and set that to zero. This gave the two equations which are correctly found and solved. – Sarvesh Ravichandran Iyer Sep 30 '21 at 10:39
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    It works if the original equation is true for all $x$. So it has to be true when $x=0$ and when $x=1$, giving you two equations in $a$ and $b$: subtract one from the other and you get "equating coefficient" of $x$, while use the $x=0$ case for "equating coefficient" of $1$ – Henry Sep 30 '21 at 10:39
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    As said, this works because the original equation is true for all $x$. Try putting $x = -5$ and $x = 11$ say, to generate 2 equations for $a$ and $b$. You will find that $a=8$ and $b=16$ as before. – Paul Sep 30 '21 at 11:17

1 Answers1

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It should have been made clear that the first line is an identity, which means that it holds for every value of $x:$ $$32x+32-8ax-4a-2a+2bx+b\equiv0.\tag0$$

As every real value satisfies this equation, its LHS polynomial has more than one root, so, by the Fundamental Theorem of Algebra, it cannot be a degree-one polynomial, so it must be the zero polynomial, so all its coefficients are $0,$ giving equations $(1)$ and $(2).$ (A fuller explanation is here.)

Alternatively: differentiating the identity gives equation $(1);$ substituting in $x=0$ gives equation $(2);$ substituting in $x=1$ gives $$64-14a+3b=0;\tag3$$ solving any two of $(1), (2)$ and $(3)$ gives the required solution set.

ryang
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  • Equation (1) is effectively $d/dx$ of the original equation, which also must vanish if the equation must vanish for all $x$. Pragmatically, it's the coefficient of $x$ in the original equation. – Michael Seifert Sep 30 '21 at 12:26
  • Thank you guys, totally makes sense now. – Justin Oct 01 '21 at 10:51