5.5: Kugawanya Polynomials
- Page ID
- 176010
Mwishoni mwa sehemu hii, utaweza:
- Kugawanya monomials
- Kugawanya polynomial na monomial
- Kugawanyika polynomials kutumia mgawanyiko mrefu
- Kugawanya polynomials kutumia mgawanyiko
- Kugawanya kazi za polynomial
- Tumia theorems zilizobaki na sababu
Kabla ya kuanza, fanya jaribio hili la utayari.
Kugawanya Monomials
Sisi sasa ni ukoo na mali yote ya exponents na kutumika yao kuzidisha polynomials. Kisha, tutatumia mali hizi kugawanya monomials na polynomials.
Kupata quotient:\(54a^2b^3÷ (−6ab^5)\).
Suluhisho
Tunapogawanya monomials na variable zaidi ya moja, tunaandika sehemu moja kwa kila kutofautiana.
\(\begin{array} {ll} {} &{54a^2b^3÷(−6ab^5)} \\[5pt] {\text{Rewrite as a fraction.}} &{\dfrac{54a^2b^3}{−6ab^5}} \\[5pt] {\text{Use fraction multiplication.}} &{\dfrac{54}{−6}·\dfrac{a^2}{a}·\dfrac{b^3}{b^5}} \\[5pt] {\text{Simplify and use the Quotient Property.}} &{−9·a·\dfrac{1}{b^2}} \\[5pt] {\text{Multiply.}} &{−\dfrac{9a}{b^2}} \end{array}\)
Kupata quotient:\(−72a^7b^3÷(8a^{12}b^4)\).
- Jibu
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\(−\dfrac{9}{a^5b}\)
Kupata quotient:\(−63c^8d^3÷(7c^{12}d^2)\).
- Jibu
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\(\dfrac{−9d}{c^4}\)
Mara baada ya kuwa ukoo na mchakato na umeifanya hatua kwa hatua mara kadhaa, unaweza kuwa na uwezo wa kurahisisha sehemu katika hatua moja.
Kupata quotient:\(\dfrac{14x^7y^{12}}{21x^{11}y^6}\).
Suluhisho
Kuwa makini sana kurahisisha\(\dfrac{14}{21}\) kwa kugawa nje sababu ya kawaida, na kurahisisha vigezo kwa kutoa exponents yao.
\(\begin{array} {ll} {} &{\dfrac{14x^7y^{12}}{21x^{11}y^6}} \\ {\text{Simplify and use the Quotient Property.}} &{\dfrac{2y^6}{3x^4}} \\ \end{array}\)
Kupata quotient:\(\dfrac{28x^5y^{14}}{49x^9y^{12}}\).
- Jibu
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\(\dfrac{4y^2}{7x^4}\)
Kupata quotient:\(\dfrac{30m^5n^{11}}{48m^{10}n^{14}}\).
- Jibu
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\(\dfrac{5}{8m^5n^3}\)
Gawanya Polynomial na Monomial
Sasa kwa kuwa tunajua jinsi ya kugawanya monomial na monomial, utaratibu unaofuata ni kugawanya polynomial ya maneno mawili au zaidi na monomial. Njia tutakayotumia kugawanya polynomial na monomial inategemea mali ya kuongeza sehemu. Hivyo tutaweza kuanza na mfano kupitia sehemu Aidha. Jumla\(\dfrac{y}{5}+\dfrac{2}{5}\) simplifies kwa\(\dfrac{y+2}{5}\). Sasa tutafanya hivyo kwa reverse ili kugawanya sehemu moja katika vipande tofauti. Kwa mfano,\(\dfrac{y+2}{5}\) inaweza kuandikwa\(\dfrac{y}{5}+\dfrac{2}{5}\).
Hii ni “reverse” ya kuongeza sehemu na inasema kwamba kama a, b, na c ni namba ambapo\(c\neq 0\), basi\(\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}\). Tutatumia hii kugawanya polynomials na monomials.
Ili kugawanya polynomial na monomial, kugawanya kila neno la polynomial na monomial.
Kupata quotient:\((18x^3y−36xy^2)÷(−3xy)\).
Suluhisho
\(\begin{array} {ll} {} &{(18x^3y−36xy^2)÷(−3xy)} \\[5pt] {\text{Rewrite as a fraction.}} &{\dfrac{18x^3y−36xy^2}{−3xy}} \\[5pt] {\text{Divide each term by the divisor. Be careful with the signs!}} &{\dfrac{18x^3y}{−3xy}−\dfrac{36xy^2}{−3xy}} \\[5pt] {\text{Simplify.}} &{−6x^2+12y} \end{array}\)
Kupata quotient:\((32a^2b−16ab^2)÷(−8ab)\).
- Jibu
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\(−4a+2b\)
Kupata quotient:\((−48a^8b^4−36a^6b^5)÷(−6a^3b^3)\).
- Jibu
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\(8a^5b+6a^3b^2\)
Gawanya Polynomials Kutumia Division
Gawanya polynomial na binomial, tunafuata utaratibu sawa na mgawanyiko mrefu wa idadi. Basi hebu tuangalie kwa makini hatua tunazochukua wakati tunagawanya nambari ya tarakimu 3, 875, na namba ya tarakimu 2, 25.
Tunaangalia mgawanyiko kwa kuzidisha quotient na mgawanyiko. Ikiwa tulifanya mgawanyiko kwa usahihi, bidhaa hiyo inapaswa kuwa sawa na mgao.
\[\begin{array} {l} {35·25} \\ {875\checkmark} \\ \nonumber \end{array}\]
Sasa tutagawanya trinomial na binomial. Unaposoma kupitia mfano, angalia jinsi hatua zilivyo sawa na mfano wa namba hapo juu.
Kupata quotient:\((x^2+9x+20)÷(x+5)\).
Suluhisho
| \(\require{enclose}\) | \(\qquad (x^2+9x+20) \div (x+5)\) |
| Andika kama tatizo la mgawanyiko mrefu. Kuwa na uhakika mgao ni katika hali ya kawaida. |
\(\qquad x+5\enclose{longdiv}{ x^2+9x+20\phantom{0}} \) |
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Gawanya\(x^2\) na\(x\). Inaweza kusaidia kujiuliza, “Ninahitaji nini |
\(\qquad \begin{array}{r} {\color{red}x}\hspace{2.3em}\\[-3pt] {\color{red}x}+5\enclose{longdiv}{ {\color{red}x^2}+9x+20\phantom{0}} \end{array}\) |
| Weka jibu,\(x\), katika quotient juu ya\(x\) muda. Kuzidisha\(x\) mara\(x+5\). Line up maneno kama chini ya mgao. |
\(\qquad \begin{array}{r}x\hspace{2.3em}\\[-3pt] x+5\enclose{longdiv}{x^2+9x+20\phantom{0}}\\[-3pt] \underline{\color{red}x^2+5x}\hspace{2.4em} \end{array}\) |
| Ondoa\(x^2+5x\) kutoka\(x^2+9x\). Unaweza kupata ni rahisi kubadili ishara na kisha kuongeza. Kisha kuleta chini ya muda wa mwisho,\(20.\) |
\(\qquad \begin{array}{r}x\hspace{2.3em}\\[-3pt] x+5\enclose{longdiv}{x^2+\phantom{00}9x+20\phantom{0}}\\[-3pt] \underline{{\color{red}-}x^2+({\color{red}-}5x)}\hspace{2.1em}\\[-3pt] {\color{red}4x+20}\hspace{0.5em} \end{array}\) |
Gawanya\(4x\) na\(x\). Inaweza kusaidia kujiuliza, “ Ninahitaji nini kuzidisha\(x\) na kupata\(4x\)?” Weka jibu,\(4\), katika quotient juu ya muda wa mara kwa mara. |
\(\qquad \begin{array}{r}x+\phantom{0}{\color{red}4}\hspace{.5em}\\[-3pt] {\color{red}x}+5\enclose{longdiv}{x^2+\phantom{00}9x+20\phantom{0}}\\[-3pt] \underline{{\color{red}-}x^2+({\color{red}-}5x)}\hspace{2.1em}\\[-3pt] {\color{red}4x}+20\hspace{0.5em} \end{array}\) |
| Panua mara 4\(x+5\). |
\(\qquad \begin{array}{r}x+\phantom{0}4\hspace{.5em}\\[-3pt] x+5\enclose{longdiv}{x^2+\phantom{00}9x+20\phantom{0}}\\[-3pt] \underline{{\color{red}-}x^2+({\color{red}-}5x)}\hspace{2.1em}\\[-3pt] 4x+20\hspace{0.5em}\\[-3pt] \underline{ \color{red}4x+20}\hspace{.5em} \end{array}\) |
| Ondoa\(4x+20\) kutoka\(4x+20\). |
\(\qquad \begin{array}{r}x+\phantom{0}4\hspace{.5em}\\[-3pt] x+5\enclose{longdiv}{x^2+\phantom{00}9x+20\phantom{0}}\\[-3pt] \underline{{\color{red}-}x^2+({\color{red}-}5x)}\hspace{2.1em}\\[-3pt] 4x+20\hspace{.5em}\\[-3pt] \underline{{\color{red}-}4x+({\color{red}-}20)}\\[-3pt] 0\hspace{.33em}\end{array}\) |
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Angalia: \(\begin{array} {ll} {\text{Multiply the quotient by the divisor.}} &{(x+4)(x+5)} \\ {\text{You should get the dividend.}} &{x^2+9x+20\checkmark}\\ \end{array}\) |
Kupata quotient:\((y^2+10y+21)÷(y+3)\).
- Jibu
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\(y+7\)
Kupata quotient:\((m^2+9m+20)÷(m+4)\).
- Jibu
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\(m+5\)
Wakati sisi kugawanywa 875 na 25, hatukuwa na salio. Lakini wakati mwingine mgawanyiko wa idadi haina kuondoka salio. Vile vile ni kweli wakati tunagawanya polynomials. Katika mfano unaofuata, tutaweza kuwa na mgawanyiko unaoacha salio. Tunaandika salio kama sehemu na mgawanyiko kama denominator.
Angalia nyuma kwenye gawio katika mifano ya awali. Masharti yaliandikwa katika utaratibu wa kushuka kwa digrii, na hapakuwa na digrii zilizopo. Mgao katika mfano huu utakuwa\(x^4−x^2+5x−6\). Inakosa\(x^3\) muda. Sisi kuongeza katika\(0x^3\) kama kishika.
Kupata quotient:\((x^4−x^2+5x−6)÷(x+2)\).
Suluhisho
Kumbuka kwamba hakuna\(x^3\) mrefu katika mgao. Sisi kuongeza\(0x^3\) kama kishika.
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| Andika kama tatizo la mgawanyiko mrefu. Kuwa na uhakika mgao ni katika hali ya kawaida na placeholders kwa masharti kukosa. |  |
| Gawanya\(x^4\) na\(x\). Weka jibu,\(x^3\), katika quotient juu ya\(x^3\) muda. Kuzidisha\(x^3\) mara\(x+2\). Weka masharti kama hayo. Ondoa na kisha kuleta chini ya muda ujao. |
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| Gawanya\(−2x^3\) na\(x\). Weka jibu,\(−2x^2\), katika quotient juu ya\(x^2\) muda. Kuzidisha\(−2x^2\) mara\(x+1\). Line up maneno kama Ondoa na kuleta chini ya muda ujao. |
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| Gawanya\(3x^2\) na\(x\). Weka jibu,\(3x\), katika quotient juu ya\(x\) muda. Kuzidisha\(3x\) mara\(x+1\). Weka masharti kama hayo. Ondoa na kuleta chini ya muda ujao. |
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| Gawanya\(−x\) na\(x\). Weka jibu,\(−1\), katika quotient juu ya muda wa mara kwa mara. Kuzidisha\(−1\) mara\(x+1\). Weka masharti kama hayo. Badilisha ishara, ongeza. Andika salio kama sehemu na mgawanyiko kama denominator. |
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| Kuangalia, kuzidisha\((x+2)(x^3−2x^2+3x−1−4x+2)\). Matokeo yanapaswa kuwa\(x^4−x^2+5x−6\). |
Kupata quotient:\((x^4−7x^2+7x+6)÷(x+3)\).
- Jibu
-
\(x^3−3x^2+2x+1+3x+3\)
Kupata quotient:\((x^4−11x^2−7x−6)÷(x+3)\).
- Jibu
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\(x^3−3x^2−2x−1−3x+3\)
Katika mfano unaofuata, tutagawanya na\(2a−3\). Tunapogawanya, tutalazimika kuzingatia vipindi pamoja na vigezo.
Kupata quotient:\((8a^3+27)÷(2a+3)\).
Suluhisho
Wakati huu tutaonyesha mgawanyiko wote kwa hatua moja. Tunahitaji kuongeza placeholders mbili ili kugawanya.
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Kuangalia, kuzidisha\((2a+3)(4a^2−6a+9)\).
Matokeo yanapaswa kuwa\(8a^3+27\).
Kupata quotient:\((x^3−64)÷(x−4)\).
- Jibu
-
\(x^2+4x+16\)
Kupata quotient:\((125x^3−8)÷(5x−2)\).
- Jibu
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\(25x^2+10x+4\)
Gawanya Polynomials kwa kutumia Division
Kama tulivyosema hapo awali, wanahisabati wanapenda kupata ruwaza ili kufanya kazi yao iwe rahisi. Kwa kuwa mgawanyiko mrefu unaweza kuwa tedious, hebu angalia nyuma katika mgawanyiko mrefu tulifanya katika Mfano na kuangalia kwa baadhi ya mifumo. Tutatumia hii kama msingi wa kile kinachoitwa mgawanyiko wa synthetic. Tatizo sawa katika muundo wa mgawanyiko wa synthetic unaonyeshwa ijayo.
Mgawanyiko wa usanifu kimsingi huondoa vigezo na namba zisizohitajika. Hapa wote\(x\) na\(x^2\) ni kuondolewa. pamoja\(−x^2\) na na\(−4x\) kama wao ni kinyume mrefu hapo juu.
- Mstari wa kwanza wa mgawanyiko wa synthetic ni coefficients ya mgao. Ya\(−5\) ni kinyume cha 5 katika mgawanyiko.
- Mstari wa pili wa mgawanyiko wa synthetic ni namba zilizoonyeshwa nyekundu katika tatizo la mgawanyiko.
- Mstari wa tatu wa mgawanyiko wa synthetic ni namba zilizoonyeshwa kwa bluu katika tatizo la mgawanyiko.
Angalia quotient na salio ni inavyoonekana katika mstari wa tatu.
\[\text{Synthetic division only works when the divisor is of the form }x−c. \nonumber \]
Mfano unaofuata utaelezea mchakato.
Tumia mgawanyiko wa synthetic ili kupata quotient na salio wakati\(2x^3+3x^2+x+8\) umegawanywa na\(x+2\).
Suluhisho
| Andika mgao na nguvu kupungua ya\(x\). |  |
| Andika coefficients ya maneno kama mstari wa kwanza wa mgawanyiko wa synthetic. |
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| Andika mgawanyiko kama\(x−c\) na mahali c katika mgawanyiko wa synthetic katika sanduku la mgawanyiko. |
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| Kuleta mgawo wa kwanza kwenye mstari wa tatu. |  |
| Panua mgawo huo kwa mgawanyiko na uweke matokeo katika mstari wa pili chini ya mgawo wa pili. |
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| Ongeza safu ya pili, kuweka matokeo katika mstari wa tatu. |  |
| Panua matokeo hayo kwa mgawanyiko na uweke matokeo katika mstari wa pili chini ya mgawo wa tatu. |
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| Ongeza safu ya tatu, kuweka matokeo katika mstari wa tatu. |  |
| Panua matokeo hayo kwa mgawanyiko na uweke matokeo katika mstari wa tatu chini ya mgawo wa tatu. |
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| Ongeza safu ya mwisho, kuweka matokeo katika mstari wa tatu. |  |
| Quotient ni\(2x^2−1x+3\) na salio ni 2. |
Mgawanyiko umekamilika. Nambari katika mstari wa tatu hutupa matokeo. Wao\(2\space\space\space−1\space\space\space3\) ni coefficients ya quotient. Quotient ni\(2x^2−1x+3\). Ya 2 katika sanduku katika mstari wa tatu ni salio.
Angalia:
\(\begin{align} (\text{quotient})(\text{divisor}) + \text{remainder} &= \text{dividend} \nonumber\\ (2x^2−1x+3)(x+2)+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3−x^2+3x+4x^2−2x+6+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3+3x^2+x+8 &= 2x^3+3x^2+x+8\checkmark \nonumber \end{align} \)
Tumia mgawanyiko wa synthetic ili kupata quotient na salio wakati\(3x^3+10x^2+6x−2\) umegawanywa na\(x+2\).
- Jibu
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\(3x^2+4x−2;\space 2\)
Tumia mgawanyiko wa synthetic ili kupata quotient na salio wakati\(4x^3+5x^2−5x+3\) umegawanywa na\(x+2\).
- Jibu
-
\(4x^2−3x+1; 1\)
Katika mfano unaofuata, tutafanya hatua zote pamoja.
Tumia mgawanyiko wa synthetic ili kupata quotient na salio wakati\(x^4−16x^2+3x+12\) umegawanywa na\(x+4\).
Suluhisho
Polynomial\(x^4−16x^2+3x+12\) ina muda wake kwa utaratibu na shahada ya kushuka lakini tunaona hakuna\(x^3\) neno. Sisi kuongeza 0 kama kishika kwa\(x^3\) muda. Kwa\(x−c\) fomu, mgawanyiko ni\(x−(−4)\).
\(4^{\text{th}}\)Tuligawanya shahada ya polynomial kwa\(1^{\text{st}}\) shahada ya polynomial hivyo quotient itakuwa\(3^{\text{rd}}\) shahada polynomial.
Kusoma kutoka mstari wa tatu, quotient ina coefficients\(1\space\space\space−4\space\space\space0\space\space\space3\), ambayo ni\(x^3−4x^2+3\). Salio
ni 0.
Tumia mgawanyiko wa synthetic ili kupata quotient na salio wakati\(x^4−16x^2+5x+20\) umegawanywa na\(x+4\).
- Jibu
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\(x^3−4x^2+5;\space 0\)
Tumia mgawanyiko wa synthetic ili kupata quotient na salio wakati\(x^4−9x^2+2x+6\) umegawanywa na\(x+3\).
- Jibu
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\(x^3−3x^2+2;\space 0\)
Gawanya Kazi za Polynomial
Kama vile polynomials inaweza kugawanywa, kazi za polynomial pia zinaweza kugawanywa.
Kwa kazi\(f(x)\) na\(g(x)\), wapi\(g(x)\neq 0\),
\[\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)} \nonumber\]
Kwa kazi\(f(x)=x^2−5x−14\) na\(g(x)=x+2\), tafuta:
- \(\left(\dfrac{f}{g}\right)(x)\)
- \(\left(\dfrac{f}{g}\right)(−4)\).
Suluhisho
ⓐ
\(\begin{array} {ll} {\text{Substitute for }f(x)\text{ and }g(x).} &{\left(\dfrac{f}{g}\right)(x)=\dfrac{x^2−5x−14}{x+2}} \\[5pt] {\text{Divide the polynomials.}} &{\left(\dfrac{f}{g}\right)(x)=x−7} \end{array} \)
ⓑ Katika sehemu ⓐ sisi kupatikana\(\left(\dfrac{f}{g}\right)(x)\) na sasa ni aliuliza kupata\(\left(\dfrac{f}{g}\right)(−4)\).
\(\begin{array} {ll} {} &{\left(\dfrac{f}{g}\right)(x)=x−7} \\[5pt] {\text{To find }\left(\dfrac{f}{g}\right)(−4), \text{ substitute }x=−4.} &{\left(\dfrac{f}{g}\right)(−4)=−4−7} \\[5pt] {} &{\left(\dfrac{f}{g}\right)(−4)=−11} \end{array}\)
Kwa kazi\(f(x)=x^2−5x−24\) na\(g(x)=x+3\), tafuta:
- \(\left(\dfrac{f}{g}\right)(x)\)
- \(\left(\dfrac{f}{g}\right)(−3)\).
- Jibu
-
\(\left(\dfrac{f}{g}\right)(x)=x−8\)
- Jibu b
-
\(\left(\dfrac{f}{g}\right)(−3)=−11\)
Kwa kazi\(f(x)=x2−5x−36\) na\(g(x)=x+4\), tafuta:
- \(\left(\dfrac{f}{g}\right)(x)\)
- \(\left(\dfrac{f}{g}\right)(−5)\).
- Jibu
-
\(\left(\dfrac{f}{g}\right)(x)=x−9\)
- Jibu b
-
\(\left(\dfrac{f}{g}\right)(x)=x−9\)
Tumia Theorem ya Salio na Sababu
Hebu tuangalie matatizo mgawanyiko tuna tu kazi kwamba kuishia na salio. Wao ni muhtasari katika chati hapa chini. Kama sisi kuchukua mgao kutoka kila tatizo mgawanyiko na kuitumia kufafanua kazi, sisi kupata kazi inavyoonekana katika chati. Wakati mgawanyiko imeandikwa kama\(x−c\), thamani ya kazi katika\(c\)\(f(c)\), ni sawa na salio kutoka tatizo mgawanyiko.
| Mgawanyo | Mgawanyiko\(x−c\) | Salio | Kazi | \(f(c)\) |
|---|---|---|---|---|
| \(x^4−x^2+5x−6\) | \ (x-c\)” data-valign="top">\(x−(−2)\) | \(−4\) | \(f(x)=x^4−x^2+5x−6\) | \ (f (c)\)” data-valign="top">\(−4\) |
| \(3x^3−2x^2−10x+8\) | \ (x-c\)” data-valign="top">\(x−2\) | 4 | \(f(x)=3x^3−2x^2−10x+8\) | \ (f (c)\)” data-valign="top">4 |
| \(x^4−16x^2+3x+15\) | \ (x-c\)” data-valign="top">\(x−(−4)\) | 3 | \(f(x)=x^4−16x^2+3x+15\) | \ (f (c)\)” data-valign="top">3 |
Kuona hili kwa ujumla zaidi, tunatambua tunaweza kuangalia tatizo la mgawanyiko kwa kuzidisha mara quotient mgawanyiko na kuongeza salio. Katika kazi nukuu tunaweza kusema, kupata mgao\(f(x)\), sisi kuzidisha quotient,\(q(x)\) mara mgawanyiko,\(x−c\), na kuongeza salio,\(r\).
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| Ikiwa tunatathmini hili\(c\), tunapata: |  |
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Hii inatuongoza kwenye Theorem ya Salio.
Ikiwa kazi ya polynomial\(f(x)\) imegawanywa na\(x−c\), basi salio ni\(f(c)\).
Matumizi Theorem Salio kupata salio wakati\(f(x)=x^3+3x+19\) imegawanywa na\(x+2\).
Suluhisho
Ili kutumia Theorem ya Salio, tunapaswa kutumia kigawanyo kwa\(x−c\) fomu. Tunaweza kuandika mgawanyiko\(x+2\) kama\(x−(−2)\). Kwa hiyo, yetu\(c\) ni\(−2\).
Ili kupata salio, tunatathmini\(f(c)\) ambayo ni\(f(−2)\).
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| Kutathmini\(f(−2)\), mbadala\(x=−2\). |  |
| Kurahisisha. |  |
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| Salio ni 5 wakati\(f(x)=x^3+3x+19\) imegawanywa na\(x+2\). | |
| Angalia: Tumia mgawanyiko wa maandishi ili uangalie. |
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| Salio ni 5. |
Matumizi Theorem Salio kupata salio wakati\(f(x)=x^3+4x+15\) imegawanywa na\(x+2\).
- Jibu
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\(−1\)
Matumizi Theorem Salio kupata salio wakati\(f(x)=x^3−7x+12\) imegawanywa na\(x+3\).
- Jibu
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\(6\)
Wakati sisi kugawanywa\(8a^3+27\) na\(2a+3\) katika Mfano matokeo yalikuwa\(4a^2−6a+9\). Kuangalia kazi yetu,\(2a+3\) tunazidisha\(4a2−6a+9\) na kupata\(8a^3+27\).
\[(4a^2−6a+9)(2a+3)=8a^3+27 \nonumber \]
Imeandikwa kwa njia hii, tunaweza kuona kwamba\(4a^2−6a+9\) na\(2a+3\) ni sababu ya\(8a^3+27\). Tulipofanya mgawanyiko, salio lilikuwa sifuri.
Wakati wowote mgawanyiko,\(x−c\), hugawanya kazi polynomial\(f(x)\),, na kusababisha salio ya sifuri, tunasema\(x−c\) ni sababu ya\(f(x)\).
Reverse pia ni kweli. Kama\(x−c\) ni sababu ya\(f(x)\) basi\(x−c\) kugawanya kazi polynomial kusababisha salio ya sifuri.
Tutasema hili katika Theorem ya Factor.
Kwa kazi yoyote ya polynomial\(f(x)\),
- ikiwa\(x−c\) ni sababu ya\(f(x)\), basi\(f(c)=0\)
- kama\(f(c)=0\), basi\(x−c\) ni sababu ya\(f(x)\)
Matumizi Salio Theorem kuamua kama\(x−4\) ni sababu ya\(f(x)=x^3−64\).
Suluhisho
Theorem Factor inatuambia kwamba\(x−4\) ni sababu ya\(f(x)=x^3−64\) kama\(f(4)=0\).
\(\begin{array} {ll} {} &{f(x)=x^3−64} \\[5pt] {\text{To evaluate }f(4) \text{ substitute } x=4.} &{f(4)=4^3−64} \\[5pt] {\text{Simplify.}} &{f(4)=64−64} \\[5pt]{\text{Subtract.}} &{f(4)=0} \end{array}\)
Kwa kuwa\(f(4)=0, x−4\) ni sababu ya\(f(x)=x^3−64\).
Matumizi Factor Theorem kuamua kama\(x−5\) ni sababu ya\(f(x)=x^3−125\).
- Jibu
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ndiyo
Matumizi Factor Theorem kuamua kama\(x−6\) ni sababu ya\(f(x)=x^3−216\).
- Jibu
-
ndiyo
Kupata rasilimali hizi online kwa maelekezo ya ziada na mazoezi na kugawa polynomials.
- Kugawanya Polynomial na Binomial
- Idara ya Usanifu na Theorem iliyobaki
Dhana muhimu
- Idara ya Polynomial na Monomial
- Ili kugawanya polynomial na monomial, kugawanya kila neno la polynomial na monomial.
- Idara ya Kazi za Polynomial
- Kwa kazi\(f(x)\) na\(g(x)\), wapi\(g(x)\neq 0\),
\(\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}\)
- Kwa kazi\(f(x)\) na\(g(x)\), wapi\(g(x)\neq 0\),
- Theorem ya Salio
- Ikiwa kazi ya polynomial\(f(x)\) imegawanywa na\(x−c\), basi salio ni\(f(c)\).
- Theorem ya sababu: Kwa kazi yoyote ya polynomial\(f(x)\),
- ikiwa\(x−c\) ni sababu ya\(f(x)\), basi\(f(c)=0\)
- kama\(f(c)=0\), basi\(x−c\) ni sababu ya\(f(x)\)