Confusion with binomial factorization
$begingroup$
Say you have the term: $-3x^2+12$ and you want to factor it out:
$$-3(x^2 - 4)$$
$$-3((x+2)(x-2))$$
However you can also write the above as:
$$-3(x+2) (x-2)$$
But in many other cases, e.g if you have $3(4 + 5)$, you can't just turn that into $3(4) + 5$ as you'll get a different result. What's the intuition behind being able to do that in the above example?
polynomials quadratics factoring
asked Dec 28 '18 at 1:16
Click UpvoteClick Upvote
1257
$endgroup$
add a comment |
$begingroup$
Say you have the term: $-3x^2+12$ and you want to factor it out:
$$-3(x^2 - 4)$$
$$-3((x+2)(x-2))$$
However you can also write the above as:
$$-3(x+2) (x-2)$$
But in many other cases, e.g if you have $3(4 + 5)$, you can't just turn that into $3(4) + 5$ as you'll get a different result. What's the intuition behind being able to do that in the above example?
polynomials quadratics factoring
asked Dec 28 '18 at 1:16
Click UpvoteClick Upvote
1257
$endgroup$
add a comment |
$begingroup$
Say you have the term: $-3x^2+12$ and you want to factor it out:
$$-3(x^2 - 4)$$
$$-3((x+2)(x-2))$$
However you can also write the above as:
$$-3(x+2) (x-2)$$
But in many other cases, e.g if you have $3(4 + 5)$, you can't just turn that into $3(4) + 5$ as you'll get a different result. What's the intuition behind being able to do that in the above example?
polynomials quadratics factoring
asked Dec 28 '18 at 1:16
Click UpvoteClick Upvote
1257
$endgroup$
Say you have the term: $-3x^2+12$ and you want to factor it out:
$$-3(x^2 - 4)$$
$$-3((x+2)(x-2))$$
However you can also write the above as:
$$-3(x+2) (x-2)$$
But in many other cases, e.g if you have $3(4 + 5)$, you can't just turn that into $3(4) + 5$ as you'll get a different result. What's the intuition behind being able to do that in the above example?
polynomials quadratics factoring
polynomials quadratics factoring
asked Dec 28 '18 at 1:16
Click UpvoteClick Upvote
1257
asked Dec 28 '18 at 1:16
Click UpvoteClick Upvote
1257
asked Dec 28 '18 at 1:16
Click UpvoteClick Upvote
1257
asked Dec 28 '18 at 1:16
Click UpvoteClick Upvote
1257
asked Dec 28 '18 at 1:16
Click UpvoteClick Upvote
1257
1257
add a comment |
add a comment |
1 Answer
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$begingroup$
You can do the above example because $(x+2)$ and $(x-2)$ are multiplied together. If you add them together, notice that
$$-3((x+2)+(x-2))neq-3(x+2)+(x-2)$$
The general rule is that
$$a(bc)=abc$$
$$a(b+c)=ab+ac$$
answered Dec 28 '18 at 1:26
LarryLarry
2,41331129
$endgroup$
2
$begingroup$
+1 It might be useful to note that the second of the rules has a name: the distributive law - and that factoring is (in a sense) undoing a distributed calculation.
$endgroup$
– Ethan Bolker
Dec 28 '18 at 1:32
add a comment |
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$begingroup$
You can do the above example because $(x+2)$ and $(x-2)$ are multiplied together. If you add them together, notice that
$$-3((x+2)+(x-2))neq-3(x+2)+(x-2)$$
The general rule is that
$$a(bc)=abc$$
$$a(b+c)=ab+ac$$
answered Dec 28 '18 at 1:26
LarryLarry
2,41331129
$endgroup$
2
$begingroup$
+1 It might be useful to note that the second of the rules has a name: the distributive law - and that factoring is (in a sense) undoing a distributed calculation.
$endgroup$
– Ethan Bolker
Dec 28 '18 at 1:32
add a comment |
$begingroup$
You can do the above example because $(x+2)$ and $(x-2)$ are multiplied together. If you add them together, notice that
$$-3((x+2)+(x-2))neq-3(x+2)+(x-2)$$
The general rule is that
$$a(bc)=abc$$
$$a(b+c)=ab+ac$$
answered Dec 28 '18 at 1:26
LarryLarry
2,41331129
$endgroup$
2
$begingroup$
+1 It might be useful to note that the second of the rules has a name: the distributive law - and that factoring is (in a sense) undoing a distributed calculation.
$endgroup$
– Ethan Bolker
Dec 28 '18 at 1:32
add a comment |
$begingroup$
You can do the above example because $(x+2)$ and $(x-2)$ are multiplied together. If you add them together, notice that
$$-3((x+2)+(x-2))neq-3(x+2)+(x-2)$$
The general rule is that
$$a(bc)=abc$$
$$a(b+c)=ab+ac$$
answered Dec 28 '18 at 1:26
LarryLarry
2,41331129
$endgroup$
You can do the above example because $(x+2)$ and $(x-2)$ are multiplied together. If you add them together, notice that
$$-3((x+2)+(x-2))neq-3(x+2)+(x-2)$$
The general rule is that
$$a(bc)=abc$$
$$a(b+c)=ab+ac$$
answered Dec 28 '18 at 1:26
LarryLarry
2,41331129
answered Dec 28 '18 at 1:26
LarryLarry
2,41331129
answered Dec 28 '18 at 1:26
LarryLarry
2,41331129
answered Dec 28 '18 at 1:26
LarryLarry
2,41331129
2,41331129
2
$begingroup$
+1 It might be useful to note that the second of the rules has a name: the distributive law - and that factoring is (in a sense) undoing a distributed calculation.
$endgroup$
– Ethan Bolker
Dec 28 '18 at 1:32
add a comment |
2
$begingroup$
+1 It might be useful to note that the second of the rules has a name: the distributive law - and that factoring is (in a sense) undoing a distributed calculation.
$endgroup$
– Ethan Bolker
Dec 28 '18 at 1:32
2
2
$begingroup$
+1 It might be useful to note that the second of the rules has a name: the distributive law - and that factoring is (in a sense) undoing a distributed calculation.
$endgroup$
– Ethan Bolker
Dec 28 '18 at 1:32
$begingroup$
+1 It might be useful to note that the second of the rules has a name: the distributive law - and that factoring is (in a sense) undoing a distributed calculation.
$endgroup$
– Ethan Bolker
Dec 28 '18 at 1:32
add a comment |
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