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6.4: Factor Maalum Bidhaa

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    Malengo ya kujifunza

    Mwishoni mwa sehemu hii, utaweza:

    • Factor kamili trinomials mraba
    • Sababu tofauti ya mraba
    • Kiasi cha sababu na tofauti za cubes

    Kabla ya kuanza, fanya jaribio hili la utayari.

    1. Kurahisisha:\((3x^2)^3\).
    2. Kuzidisha:\((m+4)^2\).
    3. Kuzidisha:\((x−3)(x+3)\).

    Tumeona kwamba baadhi ya binomials na trinomials matokeo ya bidhaa maalum-squaring binomials na kuzidisha conjugates. Ikiwa unajifunza kutambua aina hizi za polynomials, unaweza kutumia mifumo maalum ya bidhaa ili kuwazingatia kwa haraka zaidi.

    Factor Perfect Square trinomials

    Baadhi ya trinomials ni mraba kamilifu. Wanatokana na kuzidisha nyakati za binomial yenyewe. Sisi squared binomial kutumia mfano Binomial Miraba katika sura ya awali.

    Katika mabano ya wazi 3x pamoja na 4 karibu mabano mraba, 3x ni na 4 ni b. kuandika kama squared pamoja 2ab pamoja b squared, sisi kupata mabano wazi 3x karibu mabano mraba pamoja 2 mara 3x 4 pamoja 4 squared. Hii ni sawa na 9 x squared pamoja 24x pamoja 16.

    The trinomial\(9x^2+24x+16\) inaitwa trinomial kamili ya mraba. Ni mraba wa binomial\(3x+4\).

    Katika sura hii, utaanza na trinomial kamili ya mraba na kuiweka katika mambo yake kuu. Unaweza kuzingatia hii trinomial kwa kutumia mbinu zilizoelezwa katika sehemu ya mwisho, kwani ni ya fomu\(ax^2+bx+c\). Lakini ikiwa unatambua kwamba maneno ya kwanza na ya mwisho ni mraba na trinomial inafaa mfano kamili wa mraba wa trinomials, utajiokoa kazi nyingi. Hapa ni muundo-reverse ya muundo binomial mraba.

    MFANO KAMILI WA MRABA WA TRINOMIALS

    Kama\(a\) na\(b\) ni idadi halisi

    \[a^2+2ab+b^2=(a+b)^2\nonumber\]

    \[a^2−2ab+b^2=(a−b)^2\nonumber\]

    Ili utumie mfano huu, unapaswa kutambua kwamba trinomial iliyotolewa inafaa. Angalia kwanza ili uone kama mgawo wa kuongoza ni mraba kamilifu,\(a^2\). Next kuangalia kwamba muda wa mwisho ni mraba kamili,\(b^2\). Kisha angalia muda wa kati-ni bidhaa,\(2ab\)? Ikiwa kila kitu kinaangalia, unaweza kuandika kwa urahisi mambo.

    Mfano\(\PageIndex{1}\): How to Factor Perfect Square Trinomials

    Sababu:\(9x^2+12x+4\).

    Jibu

    Hatua ya 1 ni kuangalia kama trinomial inafaa mfano kamili wa mraba wa trinomials, mraba pamoja na 2ab pamoja na b squared. Kwa hili tunaangalia kama muda wa kwanza ni mraba kamilifu. 9 x mraba ni mraba wa 3x. Kisha tunaangalia kama muda wa mwisho ni mraba kamilifu. 4 ni mraba wa 2. Kisha tunaangalia kama muda wa kati ni 2ab. 12 x ni mara mbili 3x 2. Kwa hiyo tuna trinomial mraba kamili.Hatua ya 2 ni kuandika hii kama mraba wa binomial. Tunaandika kama mabano ya wazi 3x pamoja na 2 mabano ya karibu ya mraba.Hatua ya 3 ni kuangalia kwa kuzidisha.

    Mfano\(\PageIndex{2}\)

    Sababu:\(4x^2+12x+9\).

    Jibu

    \((2x+3)^2\)

    Mfano\(\PageIndex{3}\)

    Sababu:\(9y^2+24y+16\).

    Jibu

    \((3y+4)^2\)

    Ishara ya muda wa kati huamua muundo gani tutakayotumia. Wakati muda wa kati ni hasi, tunatumia mfano\(a^2−2ab+b^2\), ambayo mambo ya\((a−b)^2\).

    Hatua hizi zimefupishwa hapa.

    SABABU KAMILI YA MRABA TRINOMIALS

    \(\begin{array} {lllll} \textbf{Step 1.} &\text{Does the trinomial fit the pattern?} &\quad &\hspace{7mm} a^2+2ab+b^2 &\hspace{7mm} a^2−2ab+b^2 \\ &\text{Are the first and last terms perfect squares?} &\quad & &\\ &\text{Write them as squares.} &\quad &\hspace{5mm}(a)^2\hspace{16mm} (b)^2 &\hspace{6mm}(a)^2\hspace{16mm} (b)^2 \\ &\text{Check the middle term. Is it }2ab? &\quad &\hspace{12mm} {\,}^{\searrow}{\,}_{2·a·b}{\,}^{\swarrow} &\hspace{12mm} {\,}^{\searrow}{\,}_{2·a·b}{\,}^{\swarrow} \\ \textbf{Step 2.} &\text{Write the square of the binomial.} &\quad &\hspace{13mm} (a+b)^2 &\hspace{13mm} (a−b)^2 \\ \textbf{Step 3.} &\text{Check by multiplying.} & & & \end{array}\)

    Tutafanya kazi moja sasa ambapo muda wa kati ni hasi.

    Mfano\(\PageIndex{4}\)

    Sababu:\(81y^2−72y+16\).

    Jibu

    Masharti ya kwanza na ya mwisho ni mraba. Angalia kama muda wa kati unafaa mfano wa trinomial kamili ya mraba. Muda wa kati ni hasi, hivyo mraba wa binomial itakuwa\((a−b)^2\).

      \(81 y^{2}-72 y+16\)
    Je, masharti ya kwanza na ya mwisho ni mraba kamili? .
    Angalia muda wa kati. .
    Je, ni mechi\((a−b)^2\)? Ndiyo. .
    Andika kama mraba wa binomial. \((9 y-4)^{2}\)
    Angalia kwa kuzidisha:

    \[(9y−4)^2\nonumber\]\[(9y)^2−2·9y·4+4^2\nonumber\]\[81y^2−72y+16\checkmark\nonumber\]    		  
    Mfano\(\PageIndex{5}\)

    Sababu:\(64y^2−80y+25\).

    Jibu

    \((8y−5)^2\)

    Mfano\(\PageIndex{6}\)

    Sababu:\(16z^2−72z+81\).

    Jibu

    \((4z−9)^2\)

    Mfano unaofuata utakuwa trinomial kamili ya mraba na vigezo viwili.

    Mfano\(\PageIndex{7}\)

    Sababu:\(36x^2+84xy+49y^2\).

    Jibu
      \(36 x^{2}+84 x y+49 y^{2}\)
    Mtihani kila neno ili kuthibitisha muundo. .
    Sababu. \((6 x+7 y)^{2}\)
    Angalia kwa kuzidisha.

    \[(6x+7y)^2\nonumber\]\[(6x)^2+2·6x·7y+(7y)^2\nonumber\]\[36x^2+84xy+49y^2\checkmark\nonumber\]    		  
    Mfano\(\PageIndex{8}\)

    Sababu:\(49x^2+84xy+36y^2\).

    Jibu

    \((7x+6y)^2\)

    Mfano\(\PageIndex{9}\)

    Sababu:\(64m^2+112mn+49n^2\).

    Jibu

    \((8m+7n)^2\)

    Kumbuka hatua ya kwanza katika factoring ni kuangalia kwa sababu kubwa ya kawaida. Perfect trinomials mraba inaweza kuwa GCF katika suala zote tatu na ni lazima factored nje kwanza. Na, wakati mwingine, mara GCF imechukuliwa, utatambua trinomial kamili ya mraba.

    Mfano\(\PageIndex{10}\)

    Sababu:\(100x^2y−80xy+16y\).

    Jibu
      \(100 x^{2} y-80 x y+16 y\)
    Je, kuna GCF? Ndiyo\(4y\), hivyo factor nje. \(4 y\left(25 x^{2}-20 x+4\right)\)
    Je, hii ni trinomial mraba kamili?  
    Thibitisha muundo. .
    Sababu. \(4 y(5 x-2)^{2}\)
    Kumbuka: Weka sababu 4 y katika bidhaa ya mwisho.  

    Angalia:

    \[4y(5x−2)^2\nonumber\]\[4y[(5x)2−2·5x·2+22]\nonumber\]

    \[4y(25x2−20x+4)\nonumber\]100x2y-80xy+16y\ checkmark\]

    		  
    Mfano\(\PageIndex{11}\)

    Sababu:\(8x^2y−24xy+18y\).

    Jibu

    \(2y(2x−3)^2\)

    Mfano\(\PageIndex{12}\)

    Sababu:\(27p^2q+90pq+75q\).

    Jibu

    \(3q(3p+5)^2\)

    Sababu tofauti ya Viwanja

    Bidhaa nyingine maalum uliyoyaona katika sura iliyotangulia ilikuwa mfano wa Bidhaa za Conjugates. Wewe kutumika hii kuzidisha binomials mbili kwamba walikuwa conjugates. Hapa ni mfano:

    Tuna wazi mabano 3x minus 4 karibu mabano wazi 3x pamoja 4. Hii ni ya fomu minus b, pamoja na b.Tunaandika tena kama mabano ya wazi 3x karibu mabano mraba minus 4 squared. Hapa, 3x ni na 4 ni b Hii ni sawa na 9 x squared minus 16.

    Tofauti ya mambo ya mraba kwa bidhaa za conjugates.

    TOFAUTI YA MFANO WA MRABA

    Kama\(a\) na\(b\) ni idadi halisi,

    mraba minus b squared sawa na minus b, pamoja na b Hapa, mraba bala b mraba ni tofauti ya mraba na bala b, pamoja na b ni conjugates.

    Kumbuka, “tofauti” inahusu kuondoa. Hivyo, kwa kutumia mfano huu lazima kuhakikisha una binomial ambayo mraba mbili ni kuwa suttracted.

    Mfano\(\PageIndex{13}\): How to Factor a Trinomial Using the Difference of Squares

    Sababu:\(64y^2−1\).

    Jibu

    Hatua ya 1 ni kuangalia kama binomial 64 y squared minus 1 inafaa mfano. Kwa kuwa sisi kuangalia zifuatazo: Je, hii ni tofauti? Ndiyo. Je, masharti ya kwanza na ya mwisho ni mraba kamili? Ndiyo.
    Hatua ya 2 ni kuandika maneno yote kama mraba, Kwa hiyo, tuna mabano wazi 8y karibu mabano mraba minus 1 squared.
    Hatua ya 3 ni kuandika bidhaa za conjugates 8y minus 1, 8y pamoja na 1.
    Hatua ya 4 ni kuangalia. Tunazidisha kupata binomial ya awali

    Mfano\(\PageIndex{14}\)

    Sababu:\(121m^2−1\).

    Jibu

    \((11m−1)(11m+1)\)

    Mfano\(\PageIndex{15}\)

    Sababu:\(81y^2−1\).

    Jibu

    \((9y−1)(9y+1)\)

    SABABU TOFAUTI YA MRABA.

    \(\begin{array} {llll} \textbf{Step 1.} &\text{Does the binomial fit the pattern?} &\qquad &\hspace{5mm} a^2−b^2 \\ &\text{Is this a difference?} &\qquad &\hspace{2mm} \text{____−____} \\ &\text{Are the first and last terms perfect squares?} & & \\ \textbf{Step 2.} &\text{Write them as squares.} &\qquad &\hspace{3mm} (a)^2−(b)^2 \\ \textbf{Step 3.} &\text{Write the product of conjugates.} &\qquad &(a−b)(a+b) \\ \textbf{Step 4.} &\text{Check by multiplying.} & & \end{array}\)

    Ni muhimu kukumbuka kwamba kiasi cha mraba hazizingatii katika bidhaa za binomials. Hakuna mambo ya binomial ambayo huzidisha pamoja ili kupata jumla ya mraba. Baada ya kuondoa GCF yoyote,\(a^2+b^2\) usemi ni mkuu!

    Mfano unaofuata unaonyesha vigezo katika maneno yote mawili.

    Mfano\(\PageIndex{16}\)

    Sababu:\(144x^2−49y^2\).

    Jibu

    \(\begin{array} {lll} &\quad &144x^2−49y^2 \\ \text{Is this a difference of squares? Yes.} &\quad &(12x)^2−(7y)^2 \\ \text{Factor as the product of conjugates.} &\quad &(12x−7y)(12x+7y) \\ \text{Check by multiplying.} &\quad &(12x−7y)(12x+7y) \\ \text{Check by multiplying.} &\quad & \\ &\quad & \\ &\quad & \\ \hspace{14mm} (12x−7y)(12x+7y) &\quad & \\ \hspace{21mm} 144x^2−49y^2\checkmark &\quad & \end{array}\)

    Mfano\(\PageIndex{17}\)

    Sababu:\(196m^2−25n^2\).

    Jibu

    \((14m−5n)(14m+5n)\)

    Mfano\(\PageIndex{18}\)

    Sababu:\(121p^2−9q^2\).

    Jibu

    \((11p−3q)(11p+3q)\)

    Kama siku zote, unapaswa kuangalia sababu ya kawaida kwanza wakati wowote una kujieleza kwa sababu. Wakati mwingine sababu ya kawaida inaweza “kujificha” tofauti ya mraba na huwezi kutambua mraba kamilifu mpaka utakapofanya GCF.

    Pia, ili kuzingatia kabisa binomial katika mfano unaofuata, tutazingatia tofauti ya mraba mara mbili!

    Mfano\(\PageIndex{19}\)

    Sababu:\(48x^4y^2−243y^2\).

    Jibu

    \(\begin{array} {ll} &48x^4y^2−243y^2 \\ \text{Is there a GCF? Yes, }3y^2\text{—factor it out!} &3y^2(16x^4−81) \\ \text{Is the binomial a difference of squares? Yes.} &3y^2\left((4x^2)^2−(9)^2\right) \\ \text{Factor as a product of conjugates.} &3y^2(4x^2−9)(4x^2+9) \\ \text{Notice the first binomial is also a difference of squares!} &3y^2((2x)^2−(3)^2)(4x^2+9) \\ \text{Factor it as the product of conjugates.} &3y^2(2x−3)(2x+3)(4x^2+9) \end{array}\)

    Sababu ya mwisho, jumla ya mraba, haiwezi kuhesabiwa.

    \(\begin{array} {l} \text{Check by multiplying:} \\ \hspace{10mm} 3y^2(2x−3)(2x+3)(4x^2+9) \\ \\ \\ \hspace{15mm} 3y^2(4x^2−9)(4x^2+9) \\ \hspace{20mm} 3y^2(16x^4−81) \\ \hspace{19mm} 48x^4y^2−243y^2\checkmark\end{array}\)

    Mfano\(\PageIndex{20}\)

    Sababu:\(2x^4y^2−32y^2\).

    Jibu

    \(2y^2(x−2)(x+2)(x^2+4)\)

    Mfano\(\PageIndex{21}\)

    Sababu:\(7a^4c^2−7b^4c^2\).

    Jibu

    \(7c^2(a−b)(a+b)(a^2+b^2)\)

    Mfano unaofuata una polynomial na maneno 4. Hadi sasa, wakati hii ilitokea sisi makundi maneno katika mbili na factored kutoka huko. Hapa tutaona kwamba maneno matatu ya kwanza huunda trinomial kamili ya mraba.

    Mfano\(\PageIndex{22}\)

    Sababu:\(x^2−6x+9−y^2\).

    Jibu

    Kumbuka kwamba maneno matatu ya kwanza huunda mraba kamili ya trinomial.

      \(x^{2}-6 x+9-y^{2}\)
    Sababu kwa kuunganisha masharti matatu ya kwanza. \(\underbrace{x^{2}-6 x+9} - y^{2}\)
    Tumia muundo kamili wa mraba wa trinomial. \((x-3)^{2}-y^{2}\)
    Je, hii ni tofauti ya mraba? Ndiyo.  
    Ndiyo-waandike kama mraba. .
    Sababu kama bidhaa ya conjugates. .
      \((x-3-y)(x-3+y)\)

    Unaweza kutaka kuandika upya ufumbuzi kama\((x−y−3)(x+y−3)\).

    Mfano\(\PageIndex{23}\)

    Sababu:\(x^2−10x+25−y^2\).

    Jibu

    \((x−5−y)(x−5+y)\)

    Mfano\(\PageIndex{24}\)

    Sababu:\(x^2+6x+9−4y^2\).

    Jibu

    \((x+3−2y)(x+3+2y)\)

    Kiasi cha Kiasi na Tofauti za Cubes

    Kuna mfano mwingine maalum kwa ajili ya factoring, moja ambayo hatukutumia wakati sisi kuzidisha polynomials. Hii ni mfano wa jumla na tofauti ya cubes. Tutaandika kanuni hizi kwanza na kisha tuangalie kwa kuzidisha.

    \[a^3+b^3=(a+b)(a^2−ab+b^2\nonumber\]

    \[a^3−b^3=(a−b)(a^2+ab+b^2)\nonumber\]

    Tutaangalia muundo wa kwanza na kuacha pili kwako.

      \(\color{red}(a+b) \color{black} \left(a^{2}-a b+b^{2}\right)\)
    Kusambaza. \(\color{red}a \color{black}\left(a^{2}-a b+b^{2}\right)+ \color{red}b \color{black}\left(a^{2}-a b+b^{2}\right)\)
    Kuzidisha. \(a^{3}-a^{2} b+a b^{2}+a^{2} b-a b^{2}+b^{3}\)
    Kuchanganya kama maneno. \(a^{3}+b^{3}\)
    JUMLA NA TOFAUTI YA MUUNDO WA CUBES

    \[a^3+b^3=(a+b)(a^2−ab+b^2\nonumber\]\[a^3−b^3=(a−b)(a^2+ab+b^2)\nonumber\]

    Mwelekeo wawili huonekana sawa, sivyo? Lakini angalia ishara katika mambo. Ishara ya sababu ya binomial inafanana na ishara katika binomial ya awali. Na ishara ya muda wa kati ya sababu ya trinomial ni kinyume cha ishara katika binomial ya awali. Ikiwa unatambua mfano wa ishara, inaweza kukusaidia kukariri ruwaza.

    a cubed pamoja b cubed ni wazi mabano pamoja b karibu mabano wazi mabano mraba minus ab pamoja b mraba karibu mabano. cubed minus b cubed ni wazi mabano minus karibu mabano mabano wazi mabano wazi mraba pamoja na ab b squared karibu mabano. Katika hali zote mbili, ishara ya muda wa kwanza upande wa kulia wa equation ni sawa na ishara upande wa kushoto wa equation na ishara ya muda wa pili ni kinyume cha ishara upande wa kushoto.

    Sababu ya trinomial katika jumla na tofauti ya muundo wa cubes haiwezi kuhesabiwa.

    Inasaidia sana ikiwa unajifunza kutambua cubes ya integers kutoka 1 hadi 10, kama vile umejifunza kutambua mraba. Tumeorodhesha cubes ya integers kutoka 1 hadi 10 katika Jedwali.

    n 1 2 3 4 5 6 7 8 9 10
    \(n^3\) 1 8 27 64 125 216 343 512 729 1000
    Mfano\(\PageIndex{25}\): How to Factor the Sum or Difference of Cubes

    Sababu:\(x^3+64\).

    Jibu

    Hatua ya 1 ni kuangalia kama binomial inafaa jumla au tofauti ya muundo wa cubes. Kwa hili, sisi kuangalia kama ni jumla au tofauti. x cubed plus 64 ni jumla. Halafu tunaangalia kama maneno ya kwanza na ya mwisho ni cubes kamilifu. Wao niHatua ya 2 ni kuandika upya kama cubes. Hivyo sisi kuandika upya kama x cubed pamoja 4 cubed.Hatua ya 3 ni kutumia ama jumla au tofauti ya muundo wa cubes. Kwa kuwa hii ni jumla ya cubes, tunapata mabano ya wazi x pamoja na mabano 4 ya karibu ya wazi mabano x mraba minus 4x pamoja 4 mraba.Hatua ya 4 ni kurahisisha ndani ya mabano. Tayari ni rahisiHatua ya 5 ni kuangalia kwa kuzidisha mambo.

    Mfano\(\PageIndex{26}\)

    Sababu:\(x^3+27\).

    Jibu

    \((x+3)(x^2−3x+9)\)

    Mfano\(\PageIndex{27}\)

    Sababu:\(y^3+8\).

    Jibu

    \((y+2)(y^2−2y+4)\)

    SABABU JUMLA AU TOFAUTI YA CUBES.
    1. Je, binomial inafaa jumla au tofauti ya muundo wa cubes?
      Je, ni jumla au tofauti?
      Je, maneno ya kwanza na ya mwisho ni cubes kamili?
    2. Waandike kama cubes.
    3. Tumia ama jumla au tofauti ya muundo wa cubes.
    4. Kurahisisha ndani ya mabano.
    5. Angalia kwa kuzidisha mambo.
    Mfano\(\PageIndex{28}\)

    Sababu:\(27u^3−125v^3\).

    Jibu
      \(27 u^{3}-125 v^{3}\)
    Hii binomial ni tofauti.
    Masharti ya kwanza na ya mwisho ni cubes kamilifu.    		  
    Andika maneno kama cubes. .
    Tumia tofauti ya muundo wa cubes. .
    Kurahisisha. .
    Angalia kwa kuzidisha. Tutaacha hundi kwako.
    Mfano\(\PageIndex{29}\)

    Sababu:\(8x^3−27y^3\).

    Jibu

    \((2x−3y)(4x^2+6xy+9y^2)\)

    Mfano\(\PageIndex{30}\)

    Sababu:\(1000m^3−125n^3\).

    Jibu

    \((10m−5n)(100m^2+50mn+25n^2)\)

    Katika mfano unaofuata, sisi kwanza sababu nje GCF. Kisha tunaweza kutambua jumla ya cubes.

    Mfano\(\PageIndex{31}\)

    Sababu:\(6x^3y+48y^4\).

    Jibu
      \(6 x^{3} y+48 y^{4}\)
    Factor sababu ya kawaida. \(6 y\left(x^{3}+8 y^{3}\right)\)
    Hii binomial ni jumla
    Masharti ya kwanza na ya mwisho ni cubes kamilifu.    		  
    Andika maneno kama cubes. .
    Tumia jumla ya muundo wa cubes. .
    Kurahisisha. .

    Angalia:

    Kuangalia, unaweza kupata rahisi kuzidisha jumla ya mambo ya cubes kwanza, kisha kuzidisha bidhaa hiyo kwa 6y.6y. Tutaacha kuzidisha kwako.

    Mfano\(\PageIndex{32}\)

    Sababu:\(500p^3+4q^3\).

    Jibu

    \(4(5p+q)(25p^2−5pq+q^2)\)

    Mfano\(\PageIndex{33}\)

    Sababu:\(432c^3+686d^3\).

    Jibu

    \(2(6c+7d)(36c^2−42cd+49d^2)\)

    Muda wa kwanza katika mfano unaofuata ni cubed ya binomial.

    Mfano\(\PageIndex{34}\)

    Sababu:\((x+5)^3−64x^3\).

    Jibu
      \((x+5)^{3}-64 x^{3}\)
    Hii binomial ni tofauti. Masharti ya kwanza na ya
    mwisho ni cubes kamilifu.    		  
    Andika maneno kama cubes. .
    Tumia tofauti ya muundo wa cubes. .
    Kurahisisha. \((x+5-4 x)\left(x^{2}+10 x+25+4 x^{2}+20 x+16 x^{2}\right)\)
      \((-3 x+5)\left(21 x^{2}+30 x+25\right)\)
    Angalia kwa kuzidisha. Tutaacha hundi kwako.
    Mfano\(\PageIndex{35}\)

    Sababu:\((y+1)^3−27y^3\).

    Jibu

    \((−2y+1)(13y^2+5y+1)\)

    Mfano\(\PageIndex{36}\)

    Sababu:\((n+3)^3−125n^3\).

    Jibu

    \((−4n+3)(31n^2+21n+9)\)

    Kupata rasilimali hii online kwa maelekezo ya ziada na mazoezi na factoring bidhaa maalum.

    Dhana muhimu

    • Perfect Square Trinomials Pattern: Kama na b ni namba halisi,

      \[\begin{array} {l} a^2+2ab+b^2=(a+b)^2 \\ a^2−2ab+b^2=(a−b)^2\end{array} \nonumber\]

    • Jinsi ya kuzingatia trinomials mraba kamili.
      \(\begin{array} {lllll} \textbf{Step 1.} &\text{Does the trinomial fit the pattern?} &\quad &\hspace{7mm} a^2+2ab+b^2 &\hspace{7mm} a^2−2ab+b^2 \\ &\text{Are the first and last terms perfect squares?} &\quad & &\\ &\text{Write them as squares.} &\quad &\hspace{5mm}(a)^2\hspace{16mm} (b)^2 &\hspace{6mm}(a)^2\hspace{16mm} (b)^2 \\ &\text{Check the middle term. Is it }2ab? &\quad &\hspace{12mm} {\,}^{\searrow}{\,}_{2·a·b}{\,}^{\swarrow} &\hspace{12mm} {\,}^{\searrow}{\,}_{2·a·b}{\,}^{\swarrow} \\ \textbf{Step 2.} &\text{Write the square of the binomial.} &\quad &\hspace{13mm} (a+b)^2 &\hspace{13mm} (a−b)^2 \\ \textbf{Step 3.} &\text{Check by multiplying.} & & & \end{array}\)
    • Tofauti ya Mraba Pattern: Kama, ba, b ni namba halisi,
      mraba bala b squared ni bala b, pamoja na b Hapa, mraba bala b ni tofauti ya mraba na bala b, pamoja na b ni conjugates.
    • Jinsi ya kuzingatia tofauti za mraba.
      \(\begin{array} {llll} \textbf{Step 1.} &\text{Does the binomial fit the pattern?} &\qquad &\hspace{5mm} a^2−b^2 \\ &\text{Is this a difference?} &\qquad &\hspace{2mm} \text{____−____} \\ &\text{Are the first and last terms perfect squares?} & & \\ \textbf{Step 2.} &\text{Write them as squares.} &\qquad &\hspace{3mm} (a)^2−(b)^2 \\ \textbf{Step 3.} &\text{Write the product of conjugates.} &\qquad &(a−b)(a+b) \\ \textbf{Step 4.} &\text{Check by multiplying.} & & \end{array}\)
    • Jumla na Tofauti ya Cubes Pattern
      \(\begin{array} {l} a^3+b3=(a+b)(a^2−ab+b^2) \\ a^3−b^3=(a−b)(a^2+ab+b^2) \end{array} \)
    • Jinsi ya kuzingatia jumla au tofauti ya cubes.
      1. Je, binomial inafaa jumla au tofauti ya muundo wa cubes?
        Je, ni jumla au tofauti?
        Je, maneno ya kwanza na ya mwisho ni cubes kamili?
      2. Waandike kama cubes.
      3. Tumia ama jumla au tofauti ya muundo wa cubes.
      4. Kurahisisha ndani ya mabano
      5. Angalia kwa kuzidisha mambo.