# 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.

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

##### 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:

Tofauti ya mambo ya mraba kwa bidhaa za conjugates.

##### TOFAUTI YA MFANO WA MRABA

Kama$$a$$ na$$b$$ ni idadi halisi,

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

##### 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.

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

##### 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,
• 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.