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Question Number 59070 by Tawa1 last updated on 04/May/19

Answered by tanmay last updated on 04/May/19

$$14)f\left(x\right)=q\left(x\right)(x-\alpha )$$$$x-\alpha isfactoroff\left(x\right)iff\left(\alpha \right)=0$$$$abc-(a+b+c)(bc+ca+ab)\leftarrow puta=-bhere$$$$(-b)\left(b\right)c-(-b+b+c)(bc+c\times -b+b\times -b)$$$$-b^{2}c-c(bc-bc-b^2)$$$$=-b^{2}c-{bc}^2+{bc}^2+b^{2}c$$$$=0$$$$so[(a+b)isafacor$$$$abc-(a+b+c)(bc+ca+ab)$$$$=abc-abc-a(ca+ab)-b(bc+ca+ab)-c(bc+ca+ab)$$$$=-a^2(b+c)-b^{2}c-abc-{ab}^2-{bc}^2-c^{2}a-abc$$$$=(-)[a^{2}b+a^{2}c+b^{2}c+{bc}^2+2abc+{ab}^2+{ac}^2]$$$$=(-1)[a^2(b+c)+bc(b+c)+a{(b+c)}^2]$$$$=(-b-c)[a^2+bc+ab+ac]$$$$=(-1)(b+c)\left[a(a+b)\right)+c(a+b)]$$$$=(-1)(b+c)(a+b)(a+c)$$

Commented by Tawa1 last updated on 04/May/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by tanmay last updated on 04/May/19

$$f\left(x\right)={px}^4+{qx}^3+13x^2+4x-4=0$$$$f\left(1\right)=p+q+13+4-4=0\to p+q=-13$$$$f\left(\frac{1}{2}\right)=\frac{p}{16}+\frac{q}{8}+\frac{13}{4}+\frac{4}{2}-4=0$$$$p+2q+52-32=0$$$$p+2q+20=0$$$$p+q+13=0$$$$q+7=0\to q=-7$$$$p-7+13=0p=-6$$

Commented by Tawa1 last updated on 04/May/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by tanmay last updated on 04/May/19

$${px}^4+{qx}^3-8x^2+6=q\left(x\right)(x^2-1)+2x+1$$$$so$$$${px}^4+{qx}^3-8x^2+6-2x-1isdivisibleby(x^2-1)$$$$f\left(x\right)={px}^4+{qx}^3-8x^2-2x+5$$$$f\left(1\right)=p+q-8-2+5=0$$$$p+q-5=0$$$$f(1-)=p-q-8+2+5=0$$$$p-q-1=0$$$$p+q-5=0$$$$p-q-1=0$$$$2p-6=0\to p=3$$$$3+q-5=0$$$$q=2$$$$p=3andq=2$$

Commented by Tawa1 last updated on 04/May/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{time}$$

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