Réitigh 2x^3-10x^2+11x-7=(x-4)(2x^2+hx+3)+k

$$2 x ^ { 3 } - 10 x ^ { 2 } + 11 x - 7 = ( x - 4 ) ( 2 x ^ { 2 } + h x + 3 ) + k$$

Réitigh do h. (complex solution)

$$\left\{\begin{matrix}h=-\frac{2x^{2}-8x+k-5}{x\left(x-4\right)}\text{, }&x\neq 4\text{ and }x\neq 0\\h\in \mathrm{C}\text{, }&\left(x=0\text{ or }x=4\right)\text{ and }k=5\end{matrix}\right.$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Úsáid an t-airí dáileach chun x-4 a mhéadú faoi 2x^{2}+hx+3.

$$2x^{3}-10x^{2}+11x-7=2x^{3}+hx^{2}+3x-8x^{2}-4hx-12+k$$

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

$$2x^{3}+hx^{2}+3x-8x^{2}-4hx-12+k=2x^{3}-10x^{2}+11x-7$$

Bain 2x^{3} ón dá thaobh.

$$hx^{2}+3x-8x^{2}-4hx-12+k=2x^{3}-10x^{2}+11x-7-2x^{3}$$

Comhcheangail 2x^{3} agus -2x^{3} chun 0 a fháil.

$$hx^{2}+3x-8x^{2}-4hx-12+k=-10x^{2}+11x-7$$

Bain 3x ón dá thaobh.

$$hx^{2}-8x^{2}-4hx-12+k=-10x^{2}+11x-7-3x$$

Comhcheangail 11x agus -3x chun 8x a fháil.

$$hx^{2}-8x^{2}-4hx-12+k=-10x^{2}+8x-7$$

Cuir 8x^{2} leis an dá thaobh.

$$hx^{2}-4hx-12+k=-10x^{2}+8x-7+8x^{2}$$

Comhcheangail -10x^{2} agus 8x^{2} chun -2x^{2} a fháil.

$$hx^{2}-4hx-12+k=-2x^{2}+8x-7$$

Cuir 12 leis an dá thaobh.

$$hx^{2}-4hx+k=-2x^{2}+8x-7+12$$

Suimigh -7 agus 12 chun 5 a fháil.

$$hx^{2}-4hx+k=-2x^{2}+8x+5$$

Bain k ón dá thaobh.

$$hx^{2}-4hx=-2x^{2}+8x+5-k$$

Comhcheangail na téarmaí ar fad ina bhfuil h.

$$\left(x^{2}-4x\right)h=-2x^{2}+8x+5-k$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\left(x^{2}-4x\right)h=5-k+8x-2x^{2}$$

Roinn an dá thaobh faoi x^{2}-4x.

$$\frac{\left(x^{2}-4x\right)h}{x^{2}-4x}=\frac{5-k+8x-2x^{2}}{x^{2}-4x}$$

Má roinntear é faoi x^{2}-4x cuirtear an iolrúchán faoi x^{2}-4x ar ceal.

$$h=\frac{5-k+8x-2x^{2}}{x^{2}-4x}$$

Roinn -2x^{2}+8x+5-k faoi x^{2}-4x.

$$h=\frac{5-k+8x-2x^{2}}{x\left(x-4\right)}$$

Réitigh do h.

$$\left\{\begin{matrix}h=-\frac{2x^{2}-8x+k-5}{x\left(x-4\right)}\text{, }&x\neq 4\text{ and }x\neq 0\\h\in \mathrm{R}\text{, }&\left(x=0\text{ or }x=4\right)\text{ and }k=5\end{matrix}\right.$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Úsáid an t-airí dáileach chun x-4 a mhéadú faoi 2x^{2}+hx+3.

$$2x^{3}-10x^{2}+11x-7=2x^{3}+hx^{2}+3x-8x^{2}-4hx-12+k$$

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

$$2x^{3}+hx^{2}+3x-8x^{2}-4hx-12+k=2x^{3}-10x^{2}+11x-7$$

Bain 2x^{3} ón dá thaobh.

$$hx^{2}+3x-8x^{2}-4hx-12+k=2x^{3}-10x^{2}+11x-7-2x^{3}$$

Comhcheangail 2x^{3} agus -2x^{3} chun 0 a fháil.

$$hx^{2}+3x-8x^{2}-4hx-12+k=-10x^{2}+11x-7$$

Bain 3x ón dá thaobh.

$$hx^{2}-8x^{2}-4hx-12+k=-10x^{2}+11x-7-3x$$

Comhcheangail 11x agus -3x chun 8x a fháil.

$$hx^{2}-8x^{2}-4hx-12+k=-10x^{2}+8x-7$$

Cuir 8x^{2} leis an dá thaobh.

$$hx^{2}-4hx-12+k=-10x^{2}+8x-7+8x^{2}$$

Comhcheangail -10x^{2} agus 8x^{2} chun -2x^{2} a fháil.

$$hx^{2}-4hx-12+k=-2x^{2}+8x-7$$

Cuir 12 leis an dá thaobh.

$$hx^{2}-4hx+k=-2x^{2}+8x-7+12$$

Suimigh -7 agus 12 chun 5 a fháil.

$$hx^{2}-4hx+k=-2x^{2}+8x+5$$

Bain k ón dá thaobh.

$$hx^{2}-4hx=-2x^{2}+8x+5-k$$

Comhcheangail na téarmaí ar fad ina bhfuil h.

$$\left(x^{2}-4x\right)h=-2x^{2}+8x+5-k$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\left(x^{2}-4x\right)h=5-k+8x-2x^{2}$$

Roinn an dá thaobh faoi x^{2}-4x.

$$\frac{\left(x^{2}-4x\right)h}{x^{2}-4x}=\frac{5-k+8x-2x^{2}}{x^{2}-4x}$$

Má roinntear é faoi x^{2}-4x cuirtear an iolrúchán faoi x^{2}-4x ar ceal.

$$h=\frac{5-k+8x-2x^{2}}{x^{2}-4x}$$

Roinn -2x^{2}+8x+5-k faoi x^{2}-4x.

$$h=\frac{5-k+8x-2x^{2}}{x\left(x-4\right)}$$

Réitigh do k.

$$k=5+8x+4hx-2x^{2}-hx^{2}$$