Menu Close

Given-that-the-expression-2x-3-px-2-8x-9-is-exactly-divisable-by-x-2-6x-5-find-the-value-of-p-and-q-Hence-factorise-the-expression-fully-




Question Number 85490 by oustmuchiya@gmail.com last updated on 22/Mar/20
Given that the expression 2x^3 +px^2 −8x+9 is exactly divisable by x^2 −6x+5, find the value of p and q. Hence factorise the expression fully
$${Given}\:{that}\:{the}\:{expression}\:\mathrm{2}{x}^{\mathrm{3}} +{px}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{9}\:{is}\:{exactly}\:{divisable}\:{by}\:{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{5},\:{find}\:{the}\:{value}\:{of}\:\boldsymbol{\mathrm{p}}\:{and}\:\boldsymbol{\mathrm{q}}.\:{Hence}\:{factorise}\:{the}\:{expression}\:{fully} \\ $$
Commented by john santu last updated on 22/Mar/20
nothing q in this equation.  question wrong !
$${nothing}\:{q}\:{in}\:{this}\:{equation}. \\ $$$${question}\:{wrong}\:! \\ $$
Commented by john santu last updated on 22/Mar/20
if do you mean   2x^3  +px^2 −8x+q   by remainder theorem  2x(6x−5)+p(6x−5)−8x +q   12x^2 −10x+6px−5p−8x+q  12(6x−5)+6px−18x−5p+q  72x−60+6px−18x−5p+q  (54+6p)x+q−5p−60 ≡ 0x+0  ⇒6p=−54 ⇒p=−9  ⇒q = 60+5p = 60−45=15
$${if}\:{do}\:{you}\:{mean}\: \\ $$$$\mathrm{2}{x}^{\mathrm{3}} \:+{px}^{\mathrm{2}} −\mathrm{8}{x}+{q}\: \\ $$$${by}\:{remainder}\:{theorem} \\ $$$$\mathrm{2}{x}\left(\mathrm{6}{x}−\mathrm{5}\right)+{p}\left(\mathrm{6}{x}−\mathrm{5}\right)−\mathrm{8}{x}\:+{q}\: \\ $$$$\mathrm{12}{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{6}{px}−\mathrm{5}{p}−\mathrm{8}{x}+{q} \\ $$$$\mathrm{12}\left(\mathrm{6}{x}−\mathrm{5}\right)+\mathrm{6}{px}−\mathrm{18}{x}−\mathrm{5}{p}+{q} \\ $$$$\mathrm{72}{x}−\mathrm{60}+\mathrm{6}{px}−\mathrm{18}{x}−\mathrm{5}{p}+{q} \\ $$$$\left(\mathrm{54}+\mathrm{6}{p}\right){x}+{q}−\mathrm{5}{p}−\mathrm{60}\:\equiv\:\mathrm{0}{x}+\mathrm{0} \\ $$$$\Rightarrow\mathrm{6}{p}=−\mathrm{54}\:\Rightarrow{p}=−\mathrm{9} \\ $$$$\Rightarrow{q}\:=\:\mathrm{60}+\mathrm{5}{p}\:=\:\mathrm{60}−\mathrm{45}=\mathrm{15} \\ $$
Commented by john santu last updated on 22/Mar/20
then factorise  2x^3 −9x^2 −8x+15 = (x^2 −6x+5)(2x+3)
$${then}\:{factorise} \\ $$$$\mathrm{2}{x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{15}\:=\:\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{5}\right)\left(\mathrm{2}{x}+\mathrm{3}\right) \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *