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Алгебра 1
Course: Алгебра 1 > Unit 13
Lesson 5: Factoring quadratics by groupingFactoring quadratics: leading coefficient ≠ 1
Learn how to factor quadratic expressions as the product of two linear binomials. For example, 2x²+7x+3=(2x+1)(x+3).
Бұл сабақты бастамас бұрын не білуіңіз керек
Топтастыру әдісі арқылы мүшелі көпмүшелердің ортақ көбейткішін сыртқа шығарып, көбейткіштерге жіктеуге болады. Егер бұл тақырыппен таныс болмасаңыз біздің топтастыру арқылы көбейткіштерге жіктеуге кіріспе сабағын қарай аласыз.
Жалғастырмас бұрын алғашқы коэффициенті 1-ге тең квадраттың теңдеулерді көбейткіштерге жіктеу тақырыбымен танысуға кеңес береміз.
Бұл сабақта нені үйренесің
Осы сабақта сияқты алғашқы коэффициенті -ге тең емес квадраттық теңдеулерді топтастыру әдісі арқылы көбейткіштерге жіктеуді үйренеміз.
1-мысал: өрнегін көбейткіштерге жіктеу.
Осы екі сан арқылы бастапқы теңдеудегі -мүшесін қалай бөліп жаза алатынымызды білеміз. Яғни көпмүшемізді
түріне келтіріп жаза аламыз.
Енді көпмүшемізді топтастыру әдісі арқылы көбейткіштерге жіктей аламыз:
The factored form is .
We can check our work by showing that the factors multiply back to .
Summary
In general, we can use the following steps to factor a quadratic of the form :
- Start by finding two numbers that multiply to
and add to . - Use these numbers to split up the
-term. - Use grouping to factor the quadratic expression.
Тақырып бойынша біліміңді тексер
Example 2: Factoring
To factor , we need to find two integers with a product of and a sum of .
Since and , the numbers are and .
We can now write the term as the sum of and and use grouping to factor the polynomial:
The factored form is .
We can check our work by showing that the factors multiply back to .
Take note: In step above, notice that because the third term is negative, a "+" was inserted between the groupings to keep the expression equivalent to the original. Also, in step , we needed to factor out a negative GCF from the second grouping to reveal a common factor of . Be careful with your signs!
Тақырып бойынша біліміңді тексер
When is this method useful?
Well, clearly, the method is useful to factor quadratics of the form , even when .
However, it's not always possible to factor a quadratic expression of this form using our method.
For example, let's take the expression . To factor it, we need to find two integers with a product of and a sum of . Try as you might, you will not find two such integers.
Therefore, our method doesn't work for , and for a bunch of other quadratic expressions.
It's useful to remember, however, that if this method doesn't work, it means the expression cannot be factored as where , , , and are integers.
Why is this method working?
Let's take a deep dive into why this method is at all successful. We will have to use a bunch of letters here, but please bear with us!
Suppose the general quadratic expression can be factored as with integers , , , and .
When we expand the parentheses, we obtain the quadratic expression .
Since this expression is equivalent to , the corresponding coefficients in the two expressions must be equal! This gives us the following relationship between all the unknown letters:
Now, let's define and .
According to this definition...
, and .
And so and are the two integers we are always looking for when we use this factorization method!
The next step in the method after finding and is to split the -coefficient according to and and factor using grouping.
Indeed, if we split the -term into , we will be able to use grouping to factor our expression back into .
In conclusion, in this section we...
- started with the general expanded expression
and its general factorization , - were able to find two numbers,
and , such that and we did so by defining and , - split the
-term into , and were able to factor the expanded expression back into .
This process shows why, if an expression can indeed be factored as , our method will ensure that we find this factorization.
Thanks for pulling through!
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