7.2: Vipengele vya Quadratic vya Quadratic na Mgawo wa Uongozi 1
- Page ID
- 177537
Mwishoni mwa sehemu hii, utaweza:
- Factor trinomials ya fomu\(x^{2}+b x+c\)
- Factor trinomials ya fomu\(x^{2}+b x y+c y^{2}\)
Kabla ya kuanza, fanya jaribio hili la utayari.
- Panua: (x+4) (x+5).
Kama amekosa tatizo hili, kupitia Zoezi 6.3.31. - Kurahisisha: ⓐ -9+ (-6) ⓑ -9+6.
Kama amekosa tatizo hili, kupitia Zoezi 1.4.18. - Rahisisha: ⓐ -9 (6) ⓑ -9 (-6).
Ikiwa umekosa tatizo hili, tathmini Zoezi 1.5.1. - Kurahisisha: ⓐ |-5| ⓑ |3|.
Ikiwa umekosa tatizo hili, tathmini Zoezi 1.4.2.
Sababu Trinomials ya Fomu \(x^{2}+b x+c\)
Tayari umejifunza jinsi ya kuzidisha binomials kwa kutumia FOIL. Sasa utahitaji “kurekebisha” hii kuzidisha-kuanza na bidhaa na kuishia na mambo. Hebu tuangalie mfano wa kuzidisha binomials ili urejeshe kumbukumbu yako.
Ili kuzingatia njia ya trinomial kuanza na bidhaa,\(x^{2}+5 x+6\), na kuishia na mambo,\((x+2)(x+3)\). Unahitaji kufikiri juu ya wapi kila maneno katika trinomial yalitoka.
Neno la kwanza lilitokana na kuzidisha muda wa kwanza katika kila binomial. \(x^{2}\)Ili kupata bidhaa, kila binomial lazima kuanza na x.
\[\begin{array}{l}{x^{2}+5 x+6} \\ {(x\quad)(x\quad)}\end{array}\]
Muda wa mwisho katika trinomial ulikuja kutokana na kuzidisha muda wa mwisho katika kila binomial. Hivyo maneno ya mwisho yanapaswa kuzidi hadi 6.
Nini namba mbili zinazidisha hadi 6?
Sababu za 6 zinaweza kuwa 1 na 6, au 2 na 3. Unajuaje jozi gani ya kutumia?
Fikiria muda wa kati. Ilikuja kutokana na kuongeza maneno ya nje na ya ndani.
Hivyo namba ambazo zinapaswa kuwa na bidhaa ya 6 zitahitaji jumla ya 5. Tutajaribu uwezekano wote na muhtasari matokeo katika Jedwali \(\PageIndex{1}\)- meza itakuwa na manufaa sana wakati unafanya kazi na namba ambazo zinaweza kuhesabiwa kwa njia nyingi tofauti.
| Mambo ya 6 | Jumla ya mambo |
|---|---|
| 1,6 | \(1+6=7\) |
| 2,3 | \(2+3=5\) |
Jedwali \(\PageIndex{1}\)
Tunaona kwamba 2 na 3 ni namba zinazozidisha hadi 6 na kuongeza hadi 5. Hivyo tuna sababu ya\(x^{2}+5 x+6\). Wao ni\((x+2)(x+3)\).
\[\begin{array}{ll}{x^{2}+5 x+6} & {\text { product }} \\ {(x+2)(x+3)} & {\text { factors }}\end{array}\]
Unapaswa kuangalia hii kwa kuzidisha.
Kuangalia nyuma, sisi ilianza na\(x^{2}+5 x+6\), ambayo ni ya fomu\(x^{2}+b x+c\), ambapo b = 5 na c = 6. Tuliiweka katika binomials mbili za fomu (x+m) na (x+n).
\[\begin{array}{ll}{x^{2}+5 x+6} & {x^{2}+b x+c} \\ {(x+2)(x+3)} & {(x+m)(x+n)}\end{array}\]
Ili kupata mambo sahihi, tuligundua namba mbili m na n ambazo bidhaa zake ni c na jumla ni b.
Sababu:\(x^{2}+7 x+12\)
- Jibu
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Sababu:\(x^{2}+6 x+8\)
Sababu:\(y^{2}+8 y+15\)
Hebu tufupishe muhtasari hatua tulizotumia kupata mambo.
Factor trinomials ya fomu\(x^{2}+b x+c\).
Hatua ya 1. Andika mambo kama binomials mbili na maneno ya kwanza x:\((x \quad)(x \quad )\)
Hatua ya 2. Kupata namba mbili m na n kwamba
Kuzidisha kwa c,\(m \cdot n=c\)
Kuongeza b,\(m+n=b\)
Hatua ya 3. Tumia m na n kama masharti ya mwisho ya mambo:\((x+m)(x+n)\)
Hatua ya 4. Angalia kwa kuzidisha mambo.
Sababu:\(u^{2}+11 u+24\)
- Jibu
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Kumbuka kwamba variable ni u, hivyo sababu itakuwa na maneno ya kwanza u.
\(\begin{array}{ll} & u^{2}+11 u+24\\ {\text { Write the factors as two binomials with first terms } u \text { . }} & (u \quad)(u\quad) \\ {\text { Find two numbers that: multiply to } 24 \text { and add to } 11 .} & \end{array}\)
Mambo ya 24 Jumla ya mambo 1,24 1+24=25 2,12 2+12=14 3,8 3+8=11* 4,6 4+6=10 \(\begin{array}{ll}\text { Use } 3 \text { and } 8 \text { as the last terms of the binomials. } & (u+3)(u+8)\\ \\ \text { Check. } \\ \\ \begin{array}{l}{(u+3)(u+8)} \\ {u^{2}+3 u+8 u+24} \\ {u^{2}+11 u+24 v} \checkmark\end{array}\end{array}\)
Sababu:\(q^{2}+10 q+24\)
Sababu:\(t^{2}+14 t+24\)
Sababu:\(y^{2}+17 y+60\)
- Jibu
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\(\begin{array}{ll} & y^{2}+17 y+60\\ \text { Write the factors as two binomials with first terms y. } & (y \quad)(y\quad)\end{array}\)
Pata namba mbili zinazozidisha hadi 60 na uongeze kwenye 17.
\(\begin{array} {ll} \text { Use } 5 \text { and } 12 \text { as the last terms. } & (y+5)(y+12) \\ \text{ Check.} & \\ \\ \begin{array}{l}{(y+5)(y+12)} \\ {\left(y^{2}+12 y+5 y+60\right)} \\ {\left(y^{2}+17 y+60\right) }\checkmark \end{array} \end{array}\)Mambo ya 60 Jumla ya mambo 1,60 1+60=61 2,30 2+30=32 3,20 3+20=23 4,15 4+15=19 5,12 5+12=17* 6,10
Sababu:\(x^{2}+19 x+60\)
Sababu:\(v^{2}+23 v+60\)
Factor Trinomials ya Fomu x 2 + bx + c na b Hasi, c Chanya
Katika mifano hadi sasa, maneno yote katika trinomial yalikuwa chanya. Ni nini kinachotokea wakati kuna maneno mabaya? Naam, inategemea muda gani ni hasi. Hebu tuangalie kwanza katika trinomials na tu ya muda wa kati hasi.
Kumbuka: Ili kupata jumla hasi na bidhaa nzuri, namba lazima ziwe hasi.
Tena, fikiria juu ya FOIL na ambapo kila neno katika trinomial lilitoka. Kama vile kabla,
- neno la kwanza\(x^2\), linatokana na bidhaa za maneno mawili ya kwanza katika kila sababu ya binomial, x na y;
- chanya mwisho mrefu ni bidhaa ya masharti mawili ya mwisho
- muda wa kati hasi ni jumla ya maneno ya nje na ya ndani.
Je, unapataje bidhaa nzuri na jumla hasi? Na namba mbili hasi.
Sababu:\(t^{2}-11 t+28\)
- Jibu
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Tena, kwa kipindi chanya cha mwisho, 28, na muda wa kati hasi, -11t, tunahitaji sababu mbili hasi. Pata namba mbili zinazozidisha 28 na uongeze kwenye -11.
\(\begin{array} {ll} & t^{2}-11 t+28 \\ \text {Write the factors as two binomials with first terms } t & (t\qquad)(t\qquad)\end{array}\)
Pata namba mbili ambazo: kuzidisha hadi 28 na uongeze kwenye -11.
\(\begin{array} {ll} \text { Use }-4,-7 \text { as the last terms of the binomials. }& (t-4)(t-7) \\ \text { Check. } \\\\ \begin{array}{l}{(t-4)(t-7)} \\ {t^{2}-7 t-4 t+28} \\ {t^{2}-11 t+28}\checkmark\end{array}\end{array}\)Mambo ya 28 Jumla ya mambo -1, -28 -1+ (-28) =-29 -2, 14-14 -2+ (-14) =-16 -4, -7 \(-4+(-7)=-11^{*}\)
Sababu:\(u^{2}-9 u+18\)
Sababu:\(y^{2}-16 y+63\)
Factor Trinomials ya Fomu x2+bx+c na c Hasi
Sasa, vipi ikiwa muda wa mwisho katika trinomial ni hasi? Fikiria kuhusu FOIL. Muda wa mwisho ni bidhaa ya maneno ya mwisho katika binomials mbili. Matokeo ya bidhaa hasi kutokana na kuzidisha namba mbili na ishara tofauti. Unapaswa kuwa makini sana kuchagua mambo ili kuhakikisha unapata ishara sahihi kwa muda wa kati, pia.
Kumbuka: Ili kupata bidhaa hasi, namba lazima iwe na ishara tofauti.
Sababu:\(z^{2}+4 z-5\)
- Jibu
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Ili kupata muda mbaya wa mwisho, kuzidisha moja chanya na moja hasi. Tunahitaji mambo ya -5 kwamba kuongeza kwa chanya 4.
Mambo ya -5 Jumla ya mambo 1:5 1+ (-5) =-4 —1,5 -1+5=4* Taarifa: Tuliorodhesha wote 1, -5 na -1,5 ili kuhakikisha tulipata ishara ya muda wa kati sahihi.
\(\begin{array} {ll} &z^{2}+4 z-5 \\ \text { Factors will be two binomials with first terms z. }& (z\qquad)(z\qquad)\\ \text { Use }-1,5 \text { as the last terms of the binomials. } & (z-1)(z+5)\\ \text { Check. } & \\ \\ \begin{array}{l}{(z-1)(z+5)} \\ {z^{2}+5 z-1 z-5} \\ {z^{2}+4 z-5 }\checkmark\end{array} \end{array}\)
Sababu:\(h^{2}+4 h-12\)
- Jibu
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\((h-2)(h+6)\)
Sababu:\(: 2^{2}+k-20\)
- Jibu
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\((k-4)(k+5)\)
Hebu tufanye mabadiliko madogo kwa trinomial ya mwisho na uone ni athari gani juu ya mambo.
Sababu:\(z^{2}-4 z-5\)
- Jibu
-
Wakati huu, tunahitaji mambo ya -5 ambayo yanaongeza kwa -4.
Mambo ya -5 Jumla ya mambo 1:5 1+ (-5) =-4* —1,5 -1+5=4 \(\begin{array} {ll} &z^{2}-4 z-5 \\ \text { Factors will be two binomials with first terms z. }& (z\qquad)(z\qquad)\\ \text { Use }1,-5 \text { as the last terms of the binomials. } & (z+1)(z-5)\\ \text { Check. } & \\ \\ \begin{array}{l}{(z+1)(z-5)} \\z^{2}-5 z+1 z-5 \\ z^{2}-4 z-5\checkmark\end{array} \end{array}\)
Sababu:\(x^{2}-4 x-12\)
- Jibu
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\((x+2)(x-6)\)
Sababu:\(y^{2}-y-20\)
- Jibu
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\((y+4)(y-5)\)
Sababu:\(q^{2}-2 q-15\)
- Jibu
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\(\begin{array} {ll} &q^{2}-2 q-15\\ \text { Factors will be two binomials with first terms q. }& (q\qquad)(q\qquad)\\ \text { You can use }3,-5 \text { as the last terms of the binomials. } & (q+3)(q-5)\\ \end{array}\)
Mambo ya -15 Jumla ya mambo 1, -15 1+ (-15) =-14 —1,15 -1+15=14 3, -5 3+ (-5) =-2* -3,5 \(\begin{array}{ll}\text { Check. } & \\ \\ \begin{array}{l}{(q+3)(q-5)} \\q^{2}-5 q+3 z-15 \\ q^{2}-2q-15\checkmark\end{array} \end{array}\)
Sababu:\(r^{2}-3 r-40\)
- Jibu
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\((r+5)(r-8)\)
Sababu:\(s^{2}-3 s-10\)
- Jibu
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\((s+2)(s-5)\)
Baadhi ya trinomials ni mkuu. Njia pekee ya kuwa na uhakika wa trinomial ni mkuu ni kuorodhesha uwezekano wote na kuonyesha kwamba hakuna hata mmoja wao anayefanya kazi.
Sababu:\(y^{2}-6 y+15\)
- Jibu
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\(\begin{array}{ll}&y^{2}-6 y+15 \\ \text { Factors will be two binomials with first } & (y \qquad)(y\qquad) \\\text { terms y. } \end{array}\)
Mambo ya 15 Jumla ya mambo -1, -15 -1+ (-15) =-16 -3, -5 -3+ (-5) =-8 Kama inavyoonekana katika meza, hakuna sababu yoyote inayoongeza hadi -6; kwa hiyo, maneno ni mkuu.
Sababu:\(m^{2}+4 m+18\)
- Jibu
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mkuu
Sababu:\(n^{2}-10 n+12\)
- Jibu
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mkuu
Sababu:\(2 x+x^{2}-48\)
- Jibu
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\(\begin{array}{ll}&2 x+x^{2}-48 \\ \text { First we put the terms in decreasing degree order. } & x^{2}+2 x-48 \\ \text { Factors will be two binomials with first terms } x \text { . }& (x \qquad)(x\qquad) \end{array}\)
Kama inavyoonekana katika meza, unaweza kutumia -6,8 kama masharti ya mwisho ya binomials.
\[(x-6)(x+8)\]
Mambo ya -48 Jumla ya mambo -1,48 -1+48=47 -2,24
—3,16
-4,12
-6,8-2+24=22
-3+16=13
-4+12=8 -6+8=2\(\begin{array}{l}{\text { Check. }} \\ {(x-6)(x+8)} \\ {x^{2}-6 q+8 q-48} \\ {x^{2}+2 x-48}\checkmark \end{array}\)
Sababu:\(9 m+m^{2}+18\)
- Jibu
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\((m+3)(m+6)\)
Sababu:\(-7 n+12+n^{2}\)
- Jibu
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\((n-3)(n-4)\)
Hebu tufupishe njia ambayo sisi tu maendeleo kwa sababu trinomials ya fomu\(x^{2}+b x+c\)
Tunapofanya trinomial, tunaangalia ishara za maneno yake kwanza kuamua ishara za mambo ya binomial.
\[\begin{array}{c}{x^{2}+b x+c} \\ {(x+m)(x+n)}\end{array}\]
Wakati c ni chanya, m na n zina ishara sawa.
\[\begin{array}{cc}{\text { b positive }} & {\text { b negative }} \\ {m, n \text { positive }} & {m, n \text { negative }} \\ {x^{2}+5 x+6} & {x^{2}-6 x+8} \\ {(x+2)(x+3)} & {(x-4)(x-2)} \\ {\text { same signs }} & {\text { same signs }}\end{array}\]
Wakati c ni hasi, m na n zina ishara tofauti.
\[\begin{array}{cc}{x^{2}+x-12} & {x^{2}-2 x-15} \\ {(x+4)(x-3)} & {(x-5)(x+3)} \\ {\text { opposite signs }} & {\text { opposite signs }}\end{array}\]
Kumbuka kwamba, katika kesi wakati m na n wana ishara tofauti, ishara ya moja yenye thamani kubwa kabisa inafanana na ishara ya b.
Sababu Trinomials ya Fomu x 2 + bxy + cy 2
Wakati mwingine itabidi sababu trinomials ya fomu\(x^{2}+b x y+c y^{2}\) na vigezo mbili, kama vile\(x^{2}+12 x y+36 y^{2}\). Neno la kwanza\(x^2\),, ni bidhaa ya maneno ya kwanza ya mambo ya binomial,\(x \cdot x\). Ya\(y^2\) katika muda wa mwisho ina maana kwamba maneno ya pili ya mambo ya binomial lazima kila iwe na y. Ili kupata coefficients b na c, unatumia mchakato huo uliofupishwa katika lengo la awali.
Sababu:\(x^{2}+12 x y+36 y^{2}\)
- Jibu
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\(\begin{array}{ll }&x^{2}+12 x y+36 y^{2} \\ \text { Note that the first terms are } x, \text { last terms } &\left(x_{-} y\right)\left(x_{-} y\right) \\ \text { contain } y\end{array}\)
Pata namba zinazozidisha hadi 36 na uongeze kwenye 12.
Mambo ya 36 Jumla ya mambo 1, 36 1+36=37 2, 18 2+18=20 3, 12 3+12=15 4, 9 4+9=13 6, 6 6+6=12* \(\begin{array}{ll}{\text { Use } 6 \text { and } 6 \text { as the coefficients of the last terms. }} & (x+6 y)(x+6 y)\\ {\text { Check your answer. }}\end{array}\)
\(\begin{array}{l}{(x+6 y)(x+6 y)} \\ {x^{2}+6 x y+6 x y+36 y^{2}} \\ {x^{2}+12 x y+36 y^{2}}\checkmark \end{array}\)
Sababu:\(u^{2}+11 u v+28 v^{2}\)
- Jibu
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\((u+4 v)(u+7 v)\)
Sababu:\(x^{2}+13 x y+42 y^{2}\)
- Jibu
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\((x+6 y)(x+7 y)\)
Sababu:\(r^{2}-8 r s-9 s^{2}\)
- Jibu
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Tunahitaji\(r\) katika muda wa kwanza wa kila binomial na\(s\) katika muda wa pili. Muda wa mwisho wa trinomial ni hasi, hivyo sababu lazima ziwe na ishara tofauti.
\(\begin{array}{ll }& r^{2}-8 r s-9 s^{2} \\ \text { Note that the first terms are } r, \text { last terms contain } s &\left(r_{-} s\right)\left(r_{-} s\right) \end{array}\)
Mambo ya -9 Jumla ya mambo 1,9-9 1+ (-9) =-8* —1,9 -1+9=8 3—3 3+ (—3) =0 \(\begin{array}{ll}\text { Use } 1,-9 \text { as coefficients of the last terms. }&(r+s)(r-9 s)\\ {\text { Check your answer. }}\end{array}\)
\(\begin{array}{l}{(r-9 s)(r+s)} \\ {r^{2}+r s-9 r s-9 s^{2}} \\ {r^{2}-8 r s-9 s^{2}} \checkmark \end{array}\)
Sababu:\(a^{2}-11 a b+10 b^{2}\)
- Jibu
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\((a-b)(a-10 b)\)
Sababu:\(m^{2}-13 m n+12 n^{2}\)
- Jibu
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\((m-n)(m-12 n)\)
Sababu:\(u^{2}-9 u v-12 v^{2}\)
- Jibu
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Tunahitaji u katika muda wa kwanza wa kila binomial na v katika kipindi cha pili. Muda wa mwisho wa trinomial ni hasi, hivyo sababu lazima ziwe na ishara tofauti.
\(\begin{array}{ll }& u^{2}-9 u v-12 v^{2} \\ \text { Note that the first terms are } u, \text { last terms contain } v &\left(u_{-} v\right)\left(u_{-} v\right) \end{array}\)
Pata namba zinazozidisha hadi -12 na uongeze kwenye -9.
Mambo ya -12 Jumla ya mambo 1-12 1+ (-12) =-11 —1,12 -1+12=11 2, -6 2+ (-6) =-4 -2,6 -2+6=4 3, -4 3+ (-4) =-1 -3,4 -3+4=1 Kumbuka hakuna jozi za sababu zinazotupa -9 kama jumla. The trinomial ni mkuu.
Sababu:\(x^{2}-7 x y-10 y^{2}\)
- Jibu
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mkuu
Sababu:\(p^{2}+15 p q+20 q^{2}\)
- Jibu
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mkuu
Dhana muhimu
- Factor trinomials ya fomu\(x^{2}+b x+c\)
- Andika mambo kama binomials mbili na maneno ya kwanza\(x\):\((x\qquad)(x\qquad)\)
- Kupata namba mbili\(m\) na\(n\) kwamba
Kuzidisha kwa\(c\),\(m \cdot n=c\)
Kuongeza kwa\(b\),\(m+n=b\) - Tumia\(m\) na\(n\) kama masharti ya mwisho ya mambo:\((x+m)(x+n)\).
- Angalia kwa kuzidisha mambo.