the-polynomial-4x-3-ax-2-bx-9-where-a-and-b-consant-is-denoted-by-f-x-when-f-x-is-divide-by-x-2-the-remainder-is-r-and-when-divided-by-x-3-the-remaider-is-6r-its-further-given-that-x

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Question Number 155576 by MathsFan last updated on 02/Oct/21

$$thepolynomial4x^3+{ax}^2+bx+9,$$$$where\mathit{a}and\mathit{b}consant,isdenotedby$$$$f\left(x\right).whenf\left(x\right)isdivideby(x-2)the$$$$remainderis\mathit{r}andwhendividedby$$$$(x-3)theremaideris6\mathit{r}.itsfurther$$$$giventhat(x+3)isafactoroff\left(x\right).$$$$Showthat\mathit{b}-\mathit{a}=14$$$$andhencefind\mathit{a}and\mathit{b}$$

Commented by Rasheed.Sindhi last updated on 02/Oct/21

$$Areyousuresir/madamthat$$$$b-a=14?$$

Commented by MathsFan last updated on 02/Oct/21

$$notreallysuresir.itriedbut$$$$didntgetb-a=14.thatswhyi$$$$posteditonthisplatformforfurther$$$$convictionfromothers$$

Commented by Rasheed.Sindhi last updated on 03/Oct/21

$$Ithinkb-a=-25iscorrect.$$$$(a=4,b=-21)$$

Answered by Rasheed.Sindhi last updated on 02/Oct/21

$$\{\begin{array}{l}f\left(2\right)=4{\left(2\right)}^3+a{\left(2\right)}^2+b\left(2\right)+9=r\\ f\left(3\right)=4{\left(3\right)}^3+a{\left(3\right)}^2+b\left(3\right)+9=6r\\ f(-3)=4{(-3)}^3+a{(-3)}^2+b(-3)+9=0\end{array}$$$$\{\begin{array}{l}32+4a+2b+9=r\Rightarrow 4a+2b=r-41\\ 108+9a+3b+9=6r\Rightarrow 9a+3b=6r-117\\ -108+9a-3b+9=0\Rightarrow 3a-b=33\end{array}$$$$b=3a-33$$$$4a+2b=r-41\Rightarrow 4a+2(3a-33)=r-41$$$$10a=r-41+66=r+25$$$$r=10a-25$$$$9a+3b=6r-117$$$$\Rightarrow 9a+3(3a-33)=6(10a-25)-117$$$$18a-99=60a-150-117$$$$-42a=-168\Rightarrow a=\frac{-168}{-42}=4$$$$b=3a-33=3\left(4\right)-33=-21$$$$b=-21$$$$b-a=-21-4=-25$$

Commented by MathsFan last updated on 02/Oct/21

$$amazing$$

Answered by Rasheed.Sindhi last updated on 02/Oct/21

$$\overline{)\begin{array}{ccccc}2)& 4& a& b& 9\\ & & 8& 2a+16& 4a+2b+32\\ & 4& a+8& 2a+b+16& 4a+2b+41=r\dots A\end{array}}$$$$\overline{)\begin{array}{ccccc}3)& 4& a& b& 9\\ & & 12& 3a+36& 9a+3b+108\\ & 4& a+12& 3a+b+36& 9a+3b+117=6r\dots B\end{array}}$$$$\overline{)\begin{array}{ccccc}-3)& 4& a& b& 9\\ & & -12& -3a+36& 9a-3b-108\\ & 4& a-12& -3a+b+36& 9a-3b-99=0\dots C\end{array}}$$$$C\Rightarrow 3a-b=33\Rightarrow b=3a-33$$$$B+C:18a+18=6r\Rightarrow r=3a+3$$$$A:4a+2b+41=r$$$$4a+2(3a-33)+41=3a+3$$$$10a-66+41=3a+3$$$$7a=3+25=28$$$$a=4$$$$b=3a-33=3\left(4\right)-33=-21$$$$b-a=-21-4=-25$$

Commented by MathsFan last updated on 02/Oct/21

$$thankyousir$$$$butwhichapproachisthis$$

Commented by Rasheed.Sindhi last updated on 02/Oct/21

$$\mathcal{T}hisissyntheticdivisionapproach$$

Commented by peter frank last updated on 02/Oct/21

$$\mathrm{thanks}$$

Commented by Tawa11 last updated on 03/Oct/21

$$\mathrm{Great}\mathrm{sir}$$

Commented by Rasheed.Sindhi last updated on 03/Oct/21

$$Youarecertainlymisstawa,an$$$$oldforum-friend?$$

Answered by Rasheed.Sindhi last updated on 03/Oct/21

$$\mathrm{Verification}:$$$$a=4,b=-21$$$$f\left(x\right)=4x^3+4x^2-21x+9$$$$\bullet Whendividedbyx-2$$$$f\left(2\right)=4{\left(2\right)}^3+4{\left(2\right)}^2-21\left(2\right)+9=r$$$$=32+16-42+9=15=r$$$$\bullet Whendividedbyx-3$$$$f\left(3\right)=4{\left(3\right)}^3+4{\left(3\right)}^2-21\left(3\right)+9=6r$$$$=108+36-63+9=90=6\times 15=6r$$$$\bullet x+3isfactoroff\left(x\right)$$$$\overline{)\begin{array}{ccccc}-3)& 4& 4& -21& 9\\ & & -12& 24& -9\\ & 4& -8& 3& 0\end{array}}$$$$4x^3+4x^2-21x+9$$$$=(x+3)(4x^2-8x+3)$$$$\mathrm{V}\mathrm{e}\mathrm{r}\mathrm{i}\mathrm{f}\mathrm{i}\mathrm{e}\mathrm{d}$$$$$$

Answered by Rasheed.Sindhi last updated on 04/Oct/21

$$\mathbb{A}\mathrm{n}\mathbb{O}\mathrm{ther}\mathbb{W}\mathrm{ay}$$$${}^{\bullet }x+3\mathrm{is}\mathrm{factor}\mathrm{of}4x^3+{ax}^2+bx+9.$$$$Letotherfactoris4x^2+px+3$$$$\therefore f\left(x\right)=(x+3)(4x^2+px+3)$$$$=4x^3+(p+12)x^2+(3p+3)x+9$$$$Comparingcoefficients$$$$a=p+12\Rightarrow p=a-12,$$$$b=3p+3=3(a-12)+3=3a-33$$$$\therefore f\left(x\right)=4x^3+{ax}^2+(3a-33)x+9$$$${}^{\bullet }f\left(x\right)dividedbyx-2leaving$$$$remainderr$$$$f\left(2\right)=4{\left(2\right)}^3+a{\left(2\right)}^2+(3a-33)\left(2\right)+9=r$$$$=32+4a+6a-66+9=r$$$$10a-25=r$$$${}^{\bullet }f\left(x\right)dividedbyx-3leaving$$$$remainder6r$$$$f\left(3\right)=4{\left(3\right)}^3+a{\left(3\right)}^2+(3a-33)\left(3\right)+9=6r$$$$=108+9a+9a-99+9=6(10a-25)$$$$18a+18=60a-150$$$$42a=168\Rightarrow a=4$$$$b=3a-33=3\left(4\right)-33=-21$$$$b=-21$$$$b-a=-21-4=-25$$

Answered by ajfour last updated on 03/Oct/21

$$x=-3\Rightarrow$$$$4(-27)+9a-3b=-9$$$$\Rightarrow 9a-3b=99..\left(i\right)$$$$f\left(2\right)=r\Rightarrow$$$$32+4a+2b+9=r$$$$f\left(3\right)=6r\Rightarrow$$$$4\left(27\right)+9a+3b+9=6r$$$$3a+b+39=2r=64+8a+4b+18$$$$\Rightarrow 5a+3b=-43..\left(ii\right)$$$$\left(i\right)+\left(ii\right)\Rightarrow$$$$14a=56\Rightarrow a=4$$$$b=3a-33=-21$$$$b-a=-25$$

Answered by Rasheed.Sindhi last updated on 04/Oct/21

$$▸\begin{array}{c}f\left(3\right)=6r\\ f\left(2\right)=r\Rightarrow 6f\left(2\right)=6r\end{array}\}\Rightarrow f\left(3\right)-6f\left(2\right)=0$$$$\Rightarrow (4.3^3+a.3^2+b.3+9)$$$$-6(4.2^3+a.2^2+b.2+9)=0$$$$\Rightarrow 15a+9b=-129\dots \dots \dots \dots .A$$$$▸f(-3)=4{(-3)}^3+a{(-3)}^2+b(-3)+9=0$$$$-108+9a-3b+9=0$$$$9a-3b=99$$$$27a-9b=297\dots \dots \dots \dots .B$$$$A+B:42a=168\Rightarrow a=\frac{168}{42}=4$$$$B:15\left(4\right)+9b=-129\Rightarrow b=\frac{-129-60}{9}=-21$$$$a=4\mathrm{\&}b=-21$$$$b-a=-21-4=-25$$

Answered by Rasheed.Sindhi last updated on 04/Oct/21

$$\underset{A}{\underbrace{f\left(3\right)=6r}}\wedge \underset{B}{\underbrace{f\left(2\right)=r}}$$$$\frac{A}{B}\Rightarrow \frac{f\left(3\right)}{f\left(2\right)}=6\Rightarrow f\left(3\right)-6f\left(2\right)=0$$$$\Rightarrow 4.3^3+a.3^2+b.3+9-6(4.2^3+a.2^2+b.2+9)=0$$$$\Rightarrow 4.3^3+a.3^2+b.3+9-24.2^3-6a.2^2-6b.2-54=0$$$$\Rightarrow 108+9a+3b+9-192-24a-12b-54=0$$$$-15a-9b=129$$$$5a+3b=-43\dots \dots \dots \dots \dots \left(i\right)$$$$▸f(-3)=4{(-3)}^3+a{(-3)}^2+b(-3)+9=0$$$$\Rightarrow -108+9a-3b+9=0$$$$\Rightarrow 3a-b=33\dots \dots \dots \dots \dots .\left(ii\right)$$$$\left(i\right)\mathrm{\&}\left(ii\right):a=4,b=-21$$$$b-a=-21-4=-25$$$$$$

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