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Алгебра 1

Course: Алгебра 1 > Unit 13

Lesson 5: Factoring quadratics by grouping
Математика
Алгебра 1
Квадраттық теңдеулер: Көбейту мен көбейткіштерге жіктеу
Factoring quadratics by grouping
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Factoring quadratics: leading coefficient ≠ 1

Google Classroom
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 4 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 2, x, squared, plus, 7, x, plus, 3 сияқты алғашқы коэффициенті 1-ге тең емес квадраттық теңдеулерді топтастыру әдісі арқылы көбейткіштерге жіктеуді үйренеміз.

1-мысал: 2, x, squared, plus, 7, x, plus, 3 өрнегін көбейткіштерге жіктеу.

left parenthesis, start color #11accd, 2, end color #11accd, x, squared, start color #e07d10, plus, 7, end color #e07d10, x, start color #aa87ff, plus, 3, end color #aa87ff, right parenthesis -өрнегінің алғашқы коэффициенті start color #11accd, 2, end color #11accd-ге тең болғандықтан, осы квадраттық өрнекті көбейткіштерге жіктеуде қосылғыштарды көбейткіштерге жіктеу әдісін қолдана алмаймыз.
start color #11accd, 2, end color #11accd, x, squared, start color #e07d10, plus, 7, end color #e07d10, x, start color #aa87ff, plus, 3, end color #aa87ff өрнегін көбейткіштерге жіктеместен бұрын, көбейтіндісі start color #11accd, 2, end color #11accd, dot, start color #aa87ff, 3, end color #aa87ff, equals, 6 (алғашқы коэффициент пен тұрақты шама) сандарының көбейтіндісі сияқты және қосындысы start color #e07d10, 7, end color #e07d10-ге (x-тың коэффициенті) тең болатындай екі бүтін сан табуымыз керек.
start color #01a995, 1, end color #01a995, dot, start color #01a995, 6, end color #01a995, equals, 6 және start color #01a995, 1, end color #01a995, plus, start color #01a995, 6, end color #01a995, equals, 7 болғандықтан екі санды start color #01a995, 1, end color #01a995 және start color #01a995, 6, end color #01a995 деп ала аламыз.
Осы екі сан арқылы бастапқы теңдеудегі x-мүшесін қалай бөліп жаза алатынымызды білеміз. Яғни көпмүшемізді 2, x, squared, plus, 7, x, plus, 3, equals, 2, x, squared, plus, start color #01a995, 1, end color #01a995, x, plus, start color #01a995, 6, end color #01a995, x, plus, 3 түріне келтіріп жаза аламыз.
Енді көпмүшемізді топтастыру әдісі арқылы көбейткіштерге жіктей аламыз:
=  2x2+1x+6x+3=(2x2+1x)+(6x+3)Мүшелерді топтаймыз=x(2x+1)+3(2x+1)Ең үлкен ортақ көбейткіштерді жақшаның сыртына шығарамыз=x(2x+1)+3(2x+1)Ортақ көбейткіш!=(2x+1)(x+3) 2x+1−ді жақшаның сыртына шығарамыз\begin{aligned}&\phantom{=}~~2x^2+1x+6x+3\\\\ &=({2x^2+1x}){+(6x+3)}&&\small{\gray{\text{Мүшелерді топтаймыз}}}\\ \\ &=x({2x+1})+3({2x+1})&&\small{\gray{\text{Ең үлкен ортақ көбейткіштерді жақшаның сыртына шығарамыз}}}\\ \\ &=x(\maroonD{2x+1})+3(\maroonD{2x+1})&&\small{\gray{\text{Ортақ көбейткіш!}}}\\\\ &=(\maroonD{2x+1})(x+3)&&\small{\gray{\ 2x+1-ді \text{ жақшаның сыртына шығарамыз}}} \end{aligned}
The factored form is left parenthesis, 2, x, plus, 1, right parenthesis, left parenthesis, x, plus, 3, right parenthesis.
We can check our work by showing that the factors multiply back to 2, x, squared, plus, 7, x, plus, 3.

Summary

In general, we can use the following steps to factor a quadratic of the form start color #11accd, a, end color #11accd, x, squared, plus, start color #e07d10, b, end color #e07d10, x, plus, start color #aa87ff, c, end color #aa87ff:
  1. Start by finding two numbers that multiply to start color #11accd, a, end color #11accd, start color #aa87ff, c, end color #aa87ff and add to start color #e07d10, b, end color #e07d10.
  2. Use these numbers to split up the x-term.
  3. Use grouping to factor the quadratic expression.

Тақырып бойынша біліміңді тексер

1) Factor 3, x, squared, plus, 10, x, plus, 8.
Дұрыс жауапты таңдаңыз:

2) Factor 4, x, squared, plus, 16, x, plus, 15.

Example 2: Factoring 6, x, squared, minus, 5, x, minus, 4

To factor start color #11accd, 6, end color #11accd, x, squared, start color #e07d10, minus, 5, end color #e07d10, x, start color #aa87ff, minus, 4, end color #aa87ff, we need to find two integers with a product of start color #11accd, 6, end color #11accd, dot, left parenthesis, start color #aa87ff, minus, 4, end color #aa87ff, right parenthesis, equals, minus, 24 and a sum of start color #e07d10, minus, 5, end color #e07d10.
Since start color #01a995, 3, end color #01a995, dot, left parenthesis, start color #01a995, minus, 8, end color #01a995, right parenthesis, equals, minus, 24 and start color #01a995, 3, end color #01a995, plus, left parenthesis, start color #01a995, minus, 8, end color #01a995, right parenthesis, equals, minus, 5, the numbers are start color #01a995, 3, end color #01a995 and start color #01a995, minus, 8, end color #01a995.
We can now write the term minus, 5, x as the sum of start color #01a995, 3, end color #01a995, x and start color #01a995, minus, 8, end color #01a995, x and use grouping to factor the polynomial:
= 6x2+3x−8x−4(1)=(6x2+3x)+(−8x−4)Group terms(2)=3x(2x+1)+(−4)(2x+1)Factor out GCFs(3)=3x(2x+1)−4(2x+1)Simplify(4)=3x(2x+1)−4(2x+1)Common factor!(5)=(2x+1)(3x−4)Factor out 2x+1\begin{aligned}&&&\phantom{=}~6x^2+\tealD{3}x\tealD{-8}x-4\\\\ \small{\blueD{(1)}}&&&=({6x^2+3x}){+(-8x-4)}&&\small{\gray{\text{Group terms}}}\\ \\ \small{\blueD{(2)}}&&&=3x({2x+1})+(-4)({2x+1})&&\small{\gray{\text{Factor out GCFs}}}\\ \\ \small{\blueD{(3)}}&&&=3x({2x+1})-4({2x+1})&&\small{\gray{\text{Simplify}}}\\ \\ \small{\blueD{(4)}}&&&=3x(\maroonD{2x+1})-4(\maroonD{2x+1})&&\small{\gray{\text{Common factor!}}}\\\\ \small{\blueD{(5)}}&&&=(\maroonD{2x+1})(3x-4)&&\small{\gray{\text{Factor out } 2x+1}}\\ \end{aligned}
The factored form is left parenthesis, 2, x, plus, 1, right parenthesis, left parenthesis, 3, x, minus, 4, right parenthesis.
We can check our work by showing that the factors multiply back to 6, x, squared, minus, 5, x, minus, 4.
Take note: In step start color #11accd, left parenthesis, 1, right parenthesis, end color #11accd 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 start color #11accd, left parenthesis, 2, right parenthesis, end color #11accd, we needed to factor out a negative GCF from the second grouping to reveal a common factor of 2, x, plus, 1. Be careful with your signs!

Тақырып бойынша біліміңді тексер

3) Factor 2, x, squared, minus, 3, x, minus, 9.
Дұрыс жауапты таңдаңыз:

4) Factor 3, x, squared, minus, 2, x, minus, 5.

5) Factor 6, x, squared, minus, 13, x, plus, 6.

When is this method useful?

Well, clearly, the method is useful to factor quadratics of the form a, x, squared, plus, b, x, plus, c, even when a, does not equal, 1.
However, it's not always possible to factor a quadratic expression of this form using our method.
For example, let's take the expression start color #11accd, 2, end color #11accd, x, squared, start color #e07d10, plus, 2, end color #e07d10, x, start color #aa87ff, plus, 1, end color #aa87ff. To factor it, we need to find two integers with a product of start color #11accd, 2, end color #11accd, dot, start color #aa87ff, 1, end color #aa87ff, equals, 2 and a sum of start color #e07d10, 2, end color #e07d10. Try as you might, you will not find two such integers.
Therefore, our method doesn't work for start color #11accd, 2, end color #11accd, x, squared, start color #e07d10, plus, 2, end color #e07d10, x, start color #aa87ff, plus, 1, end color #aa87ff, 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 left parenthesis, A, x, plus, B, right parenthesis, left parenthesis, C, x, plus, D, right parenthesis where A, B, C, and D 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 a, x, squared, plus, b, x, plus, c can be factored as left parenthesis, start color #11accd, A, end color #11accd, x, plus, start color #e07d10, B, end color #e07d10, right parenthesis, left parenthesis, start color #1fab54, C, end color #1fab54, x, plus, start color #aa87ff, D, end color #aa87ff, right parenthesis with integers A, B, C, and D.
When we expand the parentheses, we obtain the quadratic expression left parenthesis, start color #11accd, A, end color #11accd, start color #1fab54, C, end color #1fab54, right parenthesis, x, squared, plus, left parenthesis, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, plus, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, right parenthesis, x, plus, start color #e07d10, B, end color #e07d10, start color #aa87ff, D, end color #aa87ff.
Since this expression is equivalent to a, x, squared, plus, b, x, plus, c, 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 m, equals, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54 and n, equals, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff.
According to this definition...
  • m, plus, n, equals, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, plus, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, equals, b, and
  • m, dot, n, equals, left parenthesis, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, right parenthesis, left parenthesis, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, right parenthesis, equals, left parenthesis, start color #11accd, A, end color #11accd, start color #1fab54, C, end color #1fab54, right parenthesis, left parenthesis, start color #e07d10, B, end color #e07d10, start color #aa87ff, D, end color #aa87ff, right parenthesis, equals, a, dot, c.
And so start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54 and start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff are the two integers we are always looking for when we use this factorization method!
The next step in the method after finding m and n is to split the x-coefficient left parenthesis, b, right parenthesis according to m and n and factor using grouping.
Indeed, if we split the x-term left parenthesis, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, plus, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, right parenthesis, x into left parenthesis, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, right parenthesis, x, plus, left parenthesis, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, right parenthesis, x, we will be able to use grouping to factor our expression back into left parenthesis, start color #11accd, A, end color #11accd, x, plus, start color #e07d10, B, end color #e07d10, right parenthesis, left parenthesis, start color #1fab54, C, end color #1fab54, x, plus, start color #aa87ff, D, end color #aa87ff, right parenthesis.
In conclusion, in this section we...
  • started with the general expanded expression a, x, squared, plus, b, x, plus, c and its general factorization left parenthesis, A, x, plus, B, right parenthesis, left parenthesis, C, x, plus, D, right parenthesis,
  • were able to find two numbers, m and n, such that m, n, equals, a, c and m, plus, n, equals, b left parenthesiswe did so by defining m, equals, B, C and n, equals, A, D, right parenthesis,
  • split the x-term b, x into m, x, plus, n, x, and were able to factor the expanded expression back into left parenthesis, A, x, plus, B, right parenthesis, left parenthesis, C, x, plus, D, right parenthesis.
This process shows why, if an expression can indeed be factored as left parenthesis, A, x, plus, B, right parenthesis, left parenthesis, C, x, plus, D, right parenthesis, our method will ensure that we find this factorization.
Thanks for pulling through!

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