# 7.3: Vipengele vya Quadratic vya Quadratic na Mgawo wa Uongozi Zaidi ya 1

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##### Malengo ya kujifunza

Mwishoni mwa sehemu hii, utaweza:

• Kutambua mkakati wa awali kwa sababu polynomials kabisa
• Factor trinomials ya fomu$$ax^{2}+bx+c$$ na GCF
• Vipengele vya trinomials kutumia jaribio na hitilafu
• Vipengele vya trinomials kutumia njia ya 'ac'
##### Kumbuka

Kabla ya kuanza, fanya jaribio hili la utayari.

1. Kupata GCF ya 45$$p^{2}$$ na 30$$p^{6}$$
Kama amekosa tatizo hili, mapitio Zoezi 7.1.4.
2. Kuzidisha$$(3 y+4)(2 y+5)$$
Ikiwa umekosa tatizo hili, tathmini Zoezi 6.3.37.
3. Kuchanganya kama maneno$$12 x^{2}+3 x+5 x+9$$
Kama amekosa tatizo hili, mapitio Zoezi 1.3.37.

## Kutambua Mkakati wa awali wa Kufanya

Hebu muhtasari ambapo sisi ni hadi sasa na factoring polynomials. Katika sehemu mbili za kwanza za sura hii, tulitumia mbinu tatu za kuzingatia: kuzingatia GCF, kuzingatia kwa makundi, na kuzingatia trinomial kwa “kufuta” FOIL. Mbinu zaidi zitafuata unapoendelea katika sura hii, na baadaye katika masomo yako ya algebra.

Je, utajuaje wakati wa kutumia kila njia ya factoring? Unapojifunza mbinu zaidi za kuzingatia, utajuaje wakati wa kutumia kila njia na usiwachanganyike? Itasaidia kuandaa mbinu za factoring katika mkakati ambao unaweza kukuongoza kutumia njia sahihi.

Unapoanza kuzingatia polynomial, daima uulize kwanza, “Je, kuna sababu kubwa ya kawaida?” Ikiwa kuna, fikiria kwanza.

Kitu kingine cha kuzingatia ni aina ya polynomial. Je, ni maneno ngapi? Je, ni binomial? Trinomial? Au ina maneno zaidi ya tatu?

• Ikiwa ni trinomial ambapo mgawo wa kuongoza ni moja$$x^{2}+b x+c$$, tumia njia ya “kufuta FOIL”.
• Ikiwa ina maneno zaidi ya tatu, jaribu njia ya kikundi. Hii ndiyo njia pekee ya kutumia kwa polynomials ya maneno zaidi ya tatu.

Baadhi ya polynomials haiwezi kufananishwa. Wanaitwa “mkuu.” Chini sisi muhtasari mbinu tuna hadi sasa.

##### CHAGUA MKAKATI WA SABABU POLYNOMIALS KABISA.
1. Je, kuna sababu kubwa ya kawaida?
• Factor ni nje.
2. Je, polynomial ni binomial, trinomial, au kuna maneno zaidi ya tatu?
• Ikiwa ni binomial, hivi sasa hatuna njia ya kuzingatia.
• Ikiwa ni trinomial ya fomu$$x^{2}+b x+c$$: Tengeneza FOIL$$(x\qquad)(x\qquad)$$
• Ikiwa ina maneno zaidi ya tatu: Tumia njia ya kikundi.
3. Angalia kwa kuzidisha mambo.

Tumia mkakati wa awali kwa sababu ya polynomial kabisa. Polynomial inachukuliwa kabisa ikiwa, isipokuwa monomials, mambo yake yote ni ya kwanza.

##### Zoezi$$\PageIndex{1}$$

Tambua njia bora ya kutumia ili kuzingatia kila polynomial.

1. $$6 y^{2}-72$$
2. $$r^{2}-10 r-24$$
3. $$p^{2}+5 p+p q+5 q$$
Jibu

$\begin{array}{ll} &6 y^{2}-72\\ \text { Is there a greatest common factor? } & \text {Yes, 6. } \\ \text { Factor out the } 6 &6\left(y^{2}-12\right) \\ \text { Is it a binomial, trinomial, or are there } & \text {Binomial, we have no method to factor } \\ \text { more than } 3 \text { terms? } & \text {binomials yet. } \end{array} \nonumber$

Jibu b

$\begin{array}{ll} &r^{2}-10 r-24\\ \text { Is there a greatest common factor? }& \text {No, there is no common factor. } \\\text { Is it a binomial, trinomial, or are there } &\text {Trinomial, with leading coefficient } 1, \text { so } \\ \text { more than three terms? }& \text {"undo" FOIL. }\end{array} \nonumber$

Jibu c

$\begin{array}{ll} &p^{2}+5 p+p q+5 q\\ \text { Is there a greatest common factor? }& \text {No, there is no common factor. } \\\text { Is it a binomial, trinomial, or are there } &\text {More than three terms, so factor using }\\ \text { more than three terms? }& \text {grouping. }\end{array} \nonumber$

##### Zoezi$$\PageIndex{2}$$

Tambua njia bora ya kutumia ili kuzingatia kila polynomial:

1. $$4 y^{2}+32$$
2. $$y^{2}+10 y+21$$
3. $$y z+2 y+3 z+6$$
Jibu

hakuna njia

Jibu b

tengua kutumia FOIL

Jibu c

sababu na kikundi

##### Zoezi$$\PageIndex{3}$$

Tambua njia bora ya kutumia ili kuzingatia kila polynomial:

1. $$a b+a+4 b+4$$
2. $$3 k^{2}+15$$
3. $$p^{2}+9 p+8$$
Jibu

sababu kwa kutumia kikundi

Jibu b

hakuna njia

Jibu c

tengua kutumia FOIL

## Factor Trinomials ya fomu shaba2 + bx + c na GCF

Sasa kwa kuwa tumeandaa kile tulichofunikwa hadi sasa, tuko tayari kuzingatia trinomials ambao mgawo wa kuongoza sio 1, trinomials ya fomu$$a x^{2}+b x+c$$. Kumbuka daima kuangalia kwa GCF kwanza! Wakati mwingine, baada ya kuzingatia GCF, mgawo wa kuongoza wa trinomial unakuwa 1 na unaweza kuifanya kwa njia katika sehemu ya mwisho. Hebu tufanye mifano michache ili kuona jinsi hii inavyofanya kazi. Jihadharini na ishara katika mifano miwili ijayo.

##### Zoezi$$\PageIndex{4}$$

Sababu kabisa:$$2 n^{2}-8 n-42$$.

Jibu

Tumia mkakati wa awali.

$$\begin{array}{ll} \text { Is there a greatest common factor? }&2 n^{2}-8 n-42\\ \text { Yes, GCF }=2 . \text { Factor it out. }& 2\left(n^{2}-4 n-21\right) \\\text { Inside the parentheses, is it a binomial, trinomial, or are there }&\\ \text { more than three terms? }& \\ \text { It is a trinomial whose coefficient is } 1, \text { so undo FOIL. } & 2(n\qquad )(n\qquad) \\ \text { Use } 3 \text { and }-7 \text { as the last terms of the binomials. } & 2(n+3)(n-7) \end{array}$$

Mambo ya -21 Jumla ya mambo
1, -21 1+ (-21) =-20
3, -7 3+ (-7) =-4*

$$\begin{array}{l}{\text {Check. }} \\ {2(n+3)(n-7)} \\ {2\left(n^{2}-7 n+3 n-21\right)} \\ {2\left(n^{2}-4 n-21\right)} \\ {2 n^{2}-8 n-42 }\checkmark \end{array}$$

##### Zoezi$$\PageIndex{5}$$

Sababu kabisa:$$4 m^{2}-4 m-8$$

Jibu

4$$(m+1)(m-2)$$

##### Zoezi$$\PageIndex{6}$$

Sababu kabisa:$$5 k^{2}-15 k-50$$

Jibu

5$$(k+2)(k-5)$$

##### Zoezi$$\PageIndex{7}$$

Sababu kabisa:$$4 y^{2}-36 y+56$$

Jibu

Tumia mkakati wa awali.
$$\begin{array}{ll} \text { Is there a greatest common factor? }&4 y^{2}-36 y+56\\ \text { Yes, GCF }=4 . \text { Factor it out. }&4\left(y^{2}-9 y+14\right) \\\text { Inside the parentheses, is it a binomial, trinomial, or are there }&\\ \text { more than three terms? }& \\ \text { It is a trinomial whose coefficient is } 1, \text { so undo FOIL. } & 4(y\qquad )(y\qquad) \\\text { Use a table like the one below to find two numbers that multiply to }&\\ 14 \text { and add to }-9\\ \text { Both factors of } 14 \text { must be negative. } & 4(y-2)(y-7) \end{array}$$

Mambo ya 14 Jumla ya mambo
-1, -14 -1+ (-14) =-15
-2, -7 -2+ (-7) =-9*

$$\begin{array}{l}{\text { Check. }} \\ {4(y-2)(y-7)} \\ {4\left(y^{2}-7 y-2 y+14\right)} \\ {4\left(y^{2}-9 y+14\right)} \\ {4 y^{2}-36 y+42 } \checkmark \end{array}$$

##### Zoezi$$\PageIndex{8}$$

Sababu kabisa:$$3 r^{2}-9 r+6$$

Jibu

3$$(r-1)(r-2)$$

##### Zoezi$$\PageIndex{9}$$

Sababu kabisa:$$2 t^{2}-10 t+12$$

Jibu

2$$(t-2)(t-3)$$

Katika mfano unaofuata GCF itajumuisha kutofautiana.

##### Zoezi$$\PageIndex{10}$$

Sababu kabisa:$$4 u^{3}+16 u^{2}-20 u$$

Jibu

Tumia mkakati wa awali.
$$\begin{array}{ll} \text { Is there a greatest common factor? }&4 u^{3}+16 u^{2}-20 u\\ \text { Yes, GCF }=4 u . \text { Factor it. }&4 u\left(u^{2}+4 u-5\right) \\\text { Binomial, trinomial, or more than three terms? }&\\ \text { more than three terms? }& \\ \text { It is a trinomial. So "undo FOIL." } & 4u(u\qquad )(u\qquad) \\\text { Use a table like the table below to find two numbers that }&4 u(u-1)(u+5)\\ \text { multiply to }-5 \text { and add to } 4\end{array}$$

Mambo ya -5 Jumla ya mambo
—1,5 -1+5=4*
1, -5 1+ (-5) =-4

Angalia.

$$\begin{array}{l}{4 u(u-1)(u+5)} \\ {4 u\left(u^{2}+5 u-u-5\right)} \\ {4 u\left(u^{2}+4 u-5\right)} \\ {4 u^{3}+16 u^{2}-20 u }\checkmark \end{array}$$

##### Zoezi$$\PageIndex{11}$$

Sababu kabisa:$$5 x^{3}+15 x^{2}-20 x$$

Jibu

5$$x(x-1)(x+4)$$

##### Zoezi$$\PageIndex{12}$$

Sababu kabisa:$$6 y^{3}+18 y^{2}-60 y$$

Jibu

6$$y(y-2)(y+5)$$

## Factor Trinomials kutumia Jaribio na Hitilafu

Ni nini kinachotokea wakati mgawo wa kuongoza sio 1 na hakuna GCF? Kuna mbinu kadhaa ambazo zinaweza kutumiwa kuzingatia hizi trinomials. Kwanza tutatumia njia ya Jaribio na Hitilafu.

Hebu tuchunguze trinomial$$3 x^{2}+5 x+2$$

Kutoka kwa kazi yetu ya awali tunatarajia hii itafanyika katika binomials mbili.

$\begin{array}{c}{3 x^{2}+5 x+2} \\ {( \qquad)( \qquad)}\end{array}$

Tunajua maneno ya kwanza ya mambo ya binomial yatazidisha kutupa 3$$x^{2}$$. Sababu pekee za 3$$x^{2}$$ ni$$1 x, 3 x$$. Tunaweza kuwaweka katika binomials.

Angalia. Je,$$1 x \cdot 3 x=3 x^{2}$$?

Tunajua masharti ya mwisho ya binomials yatazidisha hadi 2. Kwa kuwa hii trinomial ina maneno yote mazuri, tunahitaji tu kuzingatia mambo mazuri. Sababu pekee za 2 ni 1 na 2. Lakini sasa tuna kesi mbili ya kuzingatia kama itakuwa kuleta tofauti kama sisi kuandika 1, 2, au 2, 1.

Ni mambo gani sahihi? Kuamua kwamba, tunazidisha maneno ya ndani na nje.

Tangu muda wa kati wa trinomial ni 5 x, sababu katika kesi ya kwanza itafanya kazi. Hebu FOIL kuangalia.

$\begin{array}{l}{(x+1)(3 x+2)} \\ {3 x^{2}+2 x+3 x+2} \\ {3 x^{2}+5 x+2}\checkmark \end{array}$

Matokeo yetu ya factoring ni:

$\begin{array}{l}{3 x^{2}+5 x+2} \\ {(x+1)(3 x+2)}\end{array}$

##### Zoezi$$\PageIndex{13}$$: How to Factor Trinomials of the Form $$ax^2+bx+c$$ Using Trial and Error

Sababu kabisa:$$3 y^{2}+22 y+7$$

Jibu

##### Zoezi$$\PageIndex{14}$$

Sababu kabisa:$$2 a^{2}+5 a+3$$

Jibu

$$(a+1)(2 a+3)$$

##### Zoezi$$\PageIndex{15}$$

Sababu kabisa:$$4 b^{2}+5 b+1$$

Jibu

$$(b+1)(4 b+1)$$

##### VIPENGELE VYA TRINOMIALS YA FOMU$$ax^2+bx+c$$ USING TRIAL AND ERROR.
1. Andika trinomial katika utaratibu wa kushuka kwa digrii.
2. Pata jozi zote za muda wa kwanza.
3. Pata jozi zote za muda wa tatu.
4. Jaribu mchanganyiko wote unaowezekana wa mambo mpaka bidhaa sahihi inapatikana.
5. Angalia kwa kuzidisha.

Wakati muda wa kati ni hasi na muda wa mwisho ni chanya, ishara katika binomials lazima zote ziwe hasi.

##### Zoezi$$\PageIndex{16}$$

Sababu kabisa:$$6 b^{2}-13 b+5$$

Jibu
 The trinomial tayari iko katika utaratibu wa kushuka. Pata sababu za muda wa kwanza. Pata sababu za muda wa mwisho. Fikiria ishara. Tangu muda wa mwisho, 5 ni chanya, sababu zake lazima ziwe chanya au zote mbili ziwe hasi. Mgawo wa muda wa kati ni hasi, kwa hiyo tunatumia sababu hasi.
Fikiria mchanganyiko wote wa mambo.
$$6 b^{2}-13 b+5$$
Sababu zinazowezekana Bidhaa
\ (6 b^ {2} -13 b+5\) Sababu zinazowezekana” data-valign="top” class="lt-math-15168"> (b-1) (6b-5) \ (6 b^ {2} -13 b+5\) Bidhaa” data-valign="top” class="lt-math-15168">$$6 b^{2}-11 b+5$$
\ (6 b^ {2} -13 b+5\) Sababu zinazowezekana” data-valign="top” class="lt-math-15168"> (b-5) (6b-1) \ (6 b^ {2} -13 b+5\) Bidhaa” data-valign="top” class="lt-math-15168">$$6 b^{2}-31 b+5$$
\ (6 b^ {2} -13 b+5\) Sababu zinazowezekana” data-valign="top” class="lt-math-15168"> (2b-1) (3b-185) \ (6 b^ {2} -13 b+5\) Bidhaa” data-valign="top” class="lt-math-15168">$$6 b^{2}-13 b+5$$ *
\ (6 b^ {2} -13 b+5\) Sababu zinazowezekana” data-valign="top” class="lt-math-15168"> (2b-5) (3b-1) \ (6 b^ {2} -13 b+5\) Bidhaa” data-valign="top” class="lt-math-15168">$$6 b^{2}-17 b+5$$
$$\begin{array}{ll}\text{The correct factors are those whose product} & \\ \text{is the original trinomial.} & (2 b-1)(3 b-5)\\\\\text {Check by multiplying. } \\\\\begin{array}{l}{(2 b-1)(3 b-5)} \\ {6 b^{2}-10 b-3 b+5} \\ {6 b^{2}-13 b+5 v}\checkmark \end{array}\end{array}$$
##### Zoezi$$\PageIndex{17}$$

Sababu kabisa:$$8 x^{2}-14 x+3$$

Jibu

$$(2 x-3)(4 x-1)$$

##### Zoezi$$\PageIndex{18}$$

Sababu kabisa:$$10 y^{2}-37 y+7$$

Jibu

$$(2 y-7)(5 y-1)$$

Wakati sisi sababu kujieleza, sisi daima kuangalia kwa sababu kubwa ya kawaida kwanza. Ikiwa maneno hayana sababu kubwa ya kawaida, hawezi kuwa na moja katika mambo yake ama. Hii inaweza kutusaidia kuondoa baadhi ya mchanganyiko wa sababu zinazowezekana.

##### Zoezi$$\PageIndex{19}$$

Sababu kabisa:$$14 x^{2}-47 x-7$$

Jibu
 The trinomial tayari iko katika utaratibu wa kushuka. Pata sababu za muda wa kwanza. Pata sababu za muda wa mwisho. Fikiria ishara. Kwa kuwa ni hasi, jambo moja lazima liwe chanya na moja hasi.
Fikiria mchanganyiko wote wa mambo. Tunatumia kila jozi ya mambo ya 14$$x^{2}$$ na kila jozi ya mambo ya -7.
Mambo ya$$14x^2$$ Jozi na Mambo ya -7
\ (14x^2\)” data-valign="top” class="lt-math-15168">$$x, 14 x$$   11, -7
-7, 11
(utaratibu wa reverse)
\ (14x^2\)” data-valign="top” class="lt-math-15168">$$x, 14 x$$   -1, 77
77, -1
(utaratibu wa reverse)
\ (14x^2\)” data-valign="top” class="lt-math-15168">$$2x,7x$$   11, -7
-7, 11
(utaratibu wa reverse)
\ (14x^2\)” data-valign="top” class="lt-math-15168">$$2x,7x$$   -1, 77
77, -1
(utaratibu wa reverse)

Jozi hizi husababisha mchanganyiko nane wafuatayo.

$$\begin{array}{ll}\text{The correct factors are those whose product} & \\ \text{is the original trinomial.} & (2 x-7)(7 x+1)\\\\\text {Check by multiplying. } \\\\\begin{array}{l}{(2 b-1)(3 b-5)} \\ {6 b^{2}-10 b-3 b+5} \\ {6 b^{2}-13 b+5 }\checkmark \end{array}\end{array}$$
##### Zoezi$$\PageIndex{20}$$

Sababu kabisa:$$8 a^{2}-3 a-5$$

Jibu

$$(a-1)(8 a+5)$$

##### Zoezi$$\PageIndex{21}$$

Sababu kabisa:$$6 b^{2}-b-15$$

Jibu

$$(2 b+3)(3 b-5)$$

##### Zoezi$$\PageIndex{22}$$

Sababu kabisa:$$18 n^{2}-37 n+15$$

Jibu
 The trinomial tayari iko katika utaratibu wa kushuka. $$18 n^{2}-37 n+15$$ Pata sababu za muda wa kwanza. Pata sababu za muda wa mwisho. Fikiria ishara. Kwa kuwa 15 ni chanya na mgawo wa muda wa kati ni hasi, tunatumia sababu hasi.

Fikiria mchanganyiko wote wa mambo.

$$\begin{array}{ll}\text{The correct factors are those whose product} & \\ \text{is the original trinomial.} & (2 n-3)(9 n-5)\\\\\text {Check by multiplying. } \\\\ \begin{array}{l}{(2 n-3)(9 n-5)} \\ {18 n^{2}-10 n-27 n+15} \\ {18 n^{2}-37 n+15 } \checkmark\end{array} \end{array}$$
##### Zoezi$$\PageIndex{23}$$

Sababu kabisa:$$18 x^{2}-3 x-10$$

Jibu

$$(3 x+2)(6 x-5)$$

##### Zoezi$$\PageIndex{24}$$

Sababu kabisa:$$30 y^{2}-53 y-21$$

Jibu

$$(3 y+1)(10 y-21)$$

Usisahau kutafuta GCF kwanza.

##### Zoezi$$\PageIndex{25}$$

Sababu kabisa:$$10 y^{4}+55 y^{3}+60 y^{2}$$

Jibu
 $$10 y^{4}+55 y^{3}+60 y^{2}$$ Angalia kubwa ya kawaida sababu, na sababu ni ya kwanza. 5$$y^{2}\left(2 y^{2}+11 y+12\right)$$ Sababu ya trinomial.

Fikiria mchanganyiko wote.

$$\begin{array}{ll}\text{The correct factors are those whose product} &5 y^{2}(y+4)(2 y+3) \\ \text{is the original trinomial. Remember to include} & \\\text {the factor } 5 y^{2}\\\text {Check by multiplying. } \\\\ \begin{array}{l}{5 y^{2}(y+4)(2 y+3)} \\ {5 y^{2}\left(2 y^{2}+8 y+3 y+12\right)} \\ {10 y^{4}+55 y^{3}+60 y^{2}}\checkmark\end{array} \end{array}$$
##### Zoezi$$\PageIndex{26}$$

Sababu kabisa:$$15 n^{3}-85 n^{2}+100 n$$

Jibu

5$$n(n-4)(3 n-5)$$

##### Zoezi$$\PageIndex{27}$$

Sababu kabisa:$$56 q^{3}+320 q^{2}-96 q$$

Jibu

8$$q(q+6)(7 q-2)$$

## Factor Trinomials kutumia “ac” Njia

Njia nyingine ya kuzingatia trinomials ya fomu$$ax^2+bx+c$$ ni njia ya “ac”. (Njia ya “ac” wakati mwingine huitwa njia ya makundi.) Njia ya “ac” ni kweli ugani wa mbinu ulizotumia katika sehemu ya mwisho ili kuzingatia trinomials na mgawo wa kuongoza moja. Njia hii imeundwa sana (hiyo ni hatua kwa hatua), na inafanya kazi daima!

##### Zoezi$$\PageIndex{28}$$: How to Factor Trinomials Using the “ac” Method

Sababu:$$6 x^{2}+7 x+2$$

Jibu

##### Zoezi$$\PageIndex{29}$$

Sababu:$$6 x^{2}+13 x+2$$

Jibu

$$(x+2)(6 x+1)$$

##### Zoezi$$\PageIndex{30}$$

Sababu:$$4 y^{2}+8 y+3$$

Jibu

$$(2 y+1)(2 y+3)$$

##### VIPENGELE VYA TRINOMIALS YA FOMU KWA KUTUMIA NJIA YA “AC”.
1. Sababu yoyote GCF.
2. Pata ac ya bidhaa.
3. Kupata namba mbili m na n kwamba:
$$\begin{array}{ll}{\text { Multiply to } a c} & {m \cdot n=a \cdot c} \\ {\text { Add to } b} & {m+n=b}\end{array}$$
4. Split muda wa kati kwa kutumia m na n:
5. Sababu kwa kikundi.
6. Angalia kwa kuzidisha mambo.

Wakati muda wa tatu wa trinomial ni hasi, sababu za muda wa tatu zitakuwa na ishara tofauti.

##### Zoezi$$\PageIndex{31}$$

Sababu:$$8 u^{2}-17 u-21$$

Jibu
 Je, kuna sababu kubwa ya kawaida? Hapana. Kupata$$a\cdot c$$ $$a\cdot c$$ 8 (-21) -168

Pata namba mbili zinazozidisha hadi -168 na uongeze hadi -17. Sababu kubwa lazima iwe hasi.

Mambo ya -168 Jumla ya mambo
1, -168 1+ (-168) =-167
2,84 2+ (-84) =-82
3, -56 3+ (-56) =-53
4, -42 4+ (-42) =-38
6,28 6+ (-28) =-22
7,24 7+ (-24) =-17*
8,21 8+ (-21) =-13

$$\begin{array}{lc}\text { Split the middle term using } 7 u \text { and }-24 u &8 u^{2}-17 u-21 \\ & \qquad\space \swarrow\searrow \\ & \underbrace{8 u^{2}+7 u} \underbrace{-24 u-21} \\ \text { Factor by grouping. } & u(8 u+7)-3(8 u+7) \\ & (8 u+7)(u-3) \\ \text { Check by multiplying. } & \begin{array}{l}{(8 u+7)(u-3)} \\ {8 u^{2}-24 u+7 u-21} \\ {8 u^{2}-17 u-21} \checkmark \end{array} \end{array}$$

##### Zoezi$$\PageIndex{32}$$

Sababu:$$20 h^{2}+13 h-15$$

Jibu

$$(4 h-5)(5 h+3)$$

##### Zoezi$$\PageIndex{33}$$

Sababu:$$6 g^{2}+19 g-20$$

Jibu

$$(q+4)(6 q-5)$$

##### Zoezi$$\PageIndex{34}$$

Sababu:$$2 x^{2}+6 x+5$$

Jibu
 Je, kuna sababu kubwa ya kawaida? Hapana. Pata ac ac 2 (5) 10

Pata namba mbili zinazozidisha hadi 10 na uongeze kwenye 6.

Mambo ya 10 Jumla ya mambo
1,10 1+10=11
2, 5 2+5=7

Hakuna mambo ambayo huzidisha hadi 10 na kuongeza hadi 6. Polynomial ni mkuu.

##### Zoezi$$\PageIndex{35}$$

Sababu:$$10 t^{2}+19 t-15$$

Jibu

$$(2 t+5)(5 t-3)$$

##### Zoezi$$\PageIndex{36}$$

Sababu:$$3 u^{2}+8 u+5$$

Jibu

$$(u+1)(3 u+5)$$

Usisahau kuangalia kwa sababu ya kawaida!

##### Zoezi$$\PageIndex{37}$$

Sababu:$$10 y^{2}-55 y+70$$

Jibu
 Je, kuna sababu kubwa ya kawaida? Ndiyo. GCF ni 5. Factor yake. Kuwa makini kuweka sababu ya 5 njia yote kupitia suluhisho! Ya trinomial ndani ya mabano ina mgawo wa kuongoza ambao sio 1. Sababu ya trinomial. Angalia kwa kuzidisha mambo yote matatu. 5$$\left(2 y^{2}-2 y-4 y+14\right)$$ 5$$\left(2 y^{2}-11 y+14\right)$$ $$10 y^{2}-55 y+70$$✓
##### Zoezi$$\PageIndex{38}$$

Sababu:$$16 x^{2}-32 x+12$$

Jibu

4$$(2 x-3)(2 x-1)$$

##### Zoezi$$\PageIndex{39}$$

Sababu:$$18 w^{2}-39 w+18$$

Jibu

3$$(3 w-2)(2 w-3)$$

Sasa tunaweza update awali factoring Mkakati, kama inavyoonekana katika Kielelezo$$\PageIndex{1}$$ na kina katika Chagua mkakati wa sababu polynomials kabisa (updated), ni pamoja na trinomials ya fomu$$a x^{2}+b x+c$$. Kumbuka, baadhi ya polynomials ni mkuu na hivyo hawawezi kuhesabiwa.

##### CHAGUA MKAKATI WA SABABU POLYNOMIALS KABISA (UPDATED).
1. Je, kuna sababu kubwa ya kawaida?
• Factor yake.
2. Je, polynomial ni binomial, trinomial, au kuna maneno zaidi ya tatu?
• Ikiwa ni binomial, hivi sasa hatuna njia ya kuzingatia.
• Kama ni trinomial ya fomu$$x^{2}+b x+c$$
Tengeneza FOIL\$$(x\qquad)(x\qquad)$$.
• Ikiwa ni trinomial ya fomu$$a x^{2}+b x+c$$
Tumia Jaribio na Hitilafu au njia ya “ac”.
• Ikiwa ina maneno zaidi ya tatu
Tumia njia ya kikundi.
3. Angalia kwa kuzidisha mambo.
##### Kumbuka

Kupata rasilimali hizi online kwa maelekezo ya ziada na mazoezi na factoring trinomials ya fomu$$a x^{2}+b x+c$$

## Dhana muhimu

• Factor Trinomials ya Fomu kwa$$a x^{2}+b x+c$$ kutumia Jaribio na Hitilafu: Angalia Mfano.
1. Andika trinomial katika utaratibu wa kushuka kwa digrii.
2. Pata jozi zote za muda wa kwanza.
3. Pata jozi zote za muda wa tatu.
4. Jaribu mchanganyiko wote unaowezekana wa mambo mpaka bidhaa sahihi inapatikana.
5. Angalia kwa kuzidisha.
• Factor Trinomials ya Fomu$$a x^{2}+b x+c$$ Kutumia “ac” Method: Angalia Mfano.
1. Sababu yoyote GCF.
2. Pata ac ya bidhaa.
3. Kupata namba mbili m na n kwamba:$$\begin{array}{ll}{\text { Multiply to } a c} & {m \cdot n=a \cdot c} \\ {\text { Add to } b} & {m+n=b}\end{array}$$
4. Split muda wa kati kwa kutumia m na n:
5. Sababu kwa kikundi.
6. Angalia kwa kuzidisha mambo.
• Chagua mkakati wa sababu polynomials kabisa (updated):
1. Je, kuna sababu kubwa ya kawaida? Factor yake.
2. Je, polynomial ni binomial, trinomial, au kuna maneno zaidi ya tatu?
Ikiwa ni binomial, hivi sasa hatuna njia ya kuzingatia.
Ikiwa ni trinomial ya fomu$$x^2+bx+c$$
Tengeneza FOIL$$(x\qquad)(x\qquad)$$.
Ikiwa ni trinomial ya fomu$$ax^2+bx+c$$
Tumia Jaribio na Hitilafu au njia ya “ac”.
Ikiwa ina maneno zaidi ya tatu
Tumia njia ya kikundi.
3. Angalia kwa kuzidisha mambo.

## faharasa

mkuu wa polynomials
Polynomials ambayo haiwezi kuhesabiwa ni polynomials mkuu.