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Question Number 152241 by ZiYangLee last updated on 26/Aug/21

Answered by Olaf_Thorendsen last updated on 26/Aug/21

$$\left(i\right)\mathrm{and}\left(ii\right):$$$$p\left(x\right)=a(x-1)(x^2+px+q),a\ne 0$$$$\left(iii\right):$$$$p\left(0\right)=4\iff -aq=4\Rightarrow q=-\frac{4}{a}$$$$p(-1)=0\iff 2a(p-q-1)=0\Rightarrow p=\frac{a-4}{a}$$$$p\left(x\right)=a(x-1)(x^2+\frac{a-4}{a}x-\frac{4}{a})$$$$p\left(x\right)=(x-1)({ax}^2+(a-4)x-4)$$$$\left(\mathrm{iv}\right):$$$$p\left(x\right)=(x-2)\mathrm{Q}\left(x\right)+6$$$$\Rightarrow p\left(2\right)=6\iff 6a-12=6\iff a=3$$$$p\left(x\right)=(x-1)(3x^2-x-4)$$$$p\left(x\right)=3x^3-4x^2-3x+4$$

Answered by john_santu last updated on 27/Aug/21

$$\mathrm{p}\left(\mathrm{x}\right)=(\mathrm{ax}+\mathrm{b})(\mathrm{x}-1)(\mathrm{x}+1)$$$$\left(1\right)\mathrm{p}\left(0\right)=\mathrm{b}(-1)\left(1\right)=4\to \mathrm{b}=-4$$$$\left(2\right)\mathrm{p}\left(2\right)=(2\mathrm{a}-4)\left(1\right)\left(3\right)=6$$$$\to 2\mathrm{a}-4=2,\mathrm{a}=3$$$$\therefore \mathrm{p}\left(\mathrm{x}\right)=(3\mathrm{x}-4)(\mathrm{x}+1)(\mathrm{x}-1)$$$$\mathrm{p}\left(\mathrm{x}\right)=(3\mathrm{x}-4)({\mathrm{x}}^2-1)$$$$\mathrm{p}\left(\mathrm{x}\right)={3\mathrm{x}}^3-{4\mathrm{x}}^2-3\mathrm{x}+4$$

Answered by Rasheed.Sindhi last updated on 27/Aug/21

$$\left(i\right):p\left(x\right)={ax}^3+{bx}^2+cx+d$$$$\left(ii\right):p\left(1\right)=a+b+c+d=0$$$$(iii-a):p\left(0\right)=d=4\to p\left(1\right)\Rightarrow a+b+c=-4...\left(A\right)$$$$(iii-b):p(-1)=-a+b-c+4=0$$$$=-a+b-c=-4........\left(B\right)$$$$\left(A\right)+\left(B\right):2b=-8\Rightarrow b=-4$$$$\left(B\right)\Rightarrow -a-4-c=-4\Rightarrow c=-a$$$$\left(iv\right):p\left(2\right)=8a+4b+2c+d=6$$$$=8a-16+2c+4=6$$$$=4a+c=9$$$$=4a-a=9\Rightarrow a=3\Rightarrow c=-3$$$$p\left(x\right)=3x^3-4x^2-3x+4$$

Answered by Rasheed.Sindhi last updated on 27/Aug/21

$$\ll \mathcal{B}y\mathcal{S}ynthetic\mathcal{D}ivision\gg$$$$\left(i\right):{ax}^3+{bx}^2+cx+d$$$$\left(ii\right):x-1isfactorofp\left(x\right)$$$$\overline{)\begin{array}{ccccc}1)& a& b& c& d\\ & & a& a+b& a+b+c\\ & a& a+b& a+b+c& a+b+c+d=0\end{array}}$$$$\left(iiia\right):p\left(0\right)=4$$$$\overline{)\begin{array}{ccccc}0)& a& b& c& d\\ & & 0& 0& 0\\ & a& b& c& d=4\end{array}}$$$$\left(iiib\right):p(-1)=0$$$$\overline{)\begin{array}{ccccc}-1)& a& b& c& d\\ & & -a& a-b& -a+b-c\\ & a& -a+b& a-b+c& -a+b-c+d=0\end{array}}$$$$\left(iv\right):p\left(2\right)=6$$$$\overline{)\begin{array}{ccccc}2)& a& b& c& d\\ & & 2a& 4a+2b& 8a+4b+2c\\ & a& 2a+b& 4a+2b+c& 8a+4b+2c+d=6\end{array}}$$$$\{\begin{array}{l}a+b+c+d=0\\ d=4\\ -a+b-c+d=0\\ 8a+4b+2c+d=6\end{array}\Rightarrow \{\begin{array}{l}a+b+c+4=0...\left(1\right)\\ -a+b-c+4=0...\left(2\right)\\ 8a+4b+2c+4=6...\left(3\right)\end{array}$$$$\left(1\right)+\left(2\right):2b+8=0\Rightarrow b=-4$$$$\left(2\right)\Rightarrow a+c=0\Rightarrow c=-a$$$$\left(3\right)\Rightarrow 8a-16-2a+4=6\Rightarrow a=3\Rightarrow c=-3$$$$p\left(x\right)=3x^3-4x^2-3x+4$$

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