Whakaoti i te (5x+b)(cx+d)=10x^2-9x-9

$$( 5 x + b ) ( c x + d ) = 10 x ^ { 2 } - 9 x - 9$$

Whakaoti mō b

$$\left\{\begin{matrix}b=-\frac{5cx^{2}-10x^{2}+5dx+9x+9}{cx+d}\text{, }&d\neq -cx\\b\in \mathrm{R}\text{, }&\left(x=\frac{3}{2}\text{ and }c=-\frac{2d}{3}\right)\text{ or }\left(x=-\frac{3}{5}\text{ and }c=\frac{5d}{3}\right)\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te 5x+b ki te cx+d.

$$5cx^{2}+5xd+bcx+bd=10x^{2}-9x-9$$

Tangohia te 5cx^{2} mai i ngā taha e rua.

$$5xd+bcx+bd=10x^{2}-9x-9-5cx^{2}$$

Tangohia te 5xd mai i ngā taha e rua.

$$bcx+bd=10x^{2}-9x-9-5cx^{2}-5xd$$

Pahekotia ngā kīanga tau katoa e whai ana i te b.

$$\left(cx+d\right)b=10x^{2}-9x-9-5cx^{2}-5xd$$

He hanga arowhānui tō te whārite.

$$\left(cx+d\right)b=-5cx^{2}+10x^{2}-5dx-9x-9$$

Whakawehea ngā taha e rua ki te cx+d.

$$\frac{\left(cx+d\right)b}{cx+d}=\frac{-5cx^{2}+10x^{2}-5dx-9x-9}{cx+d}$$

Mā te whakawehe ki te cx+d ka wetekia te whakareanga ki te cx+d.

$$b=\frac{-5cx^{2}+10x^{2}-5dx-9x-9}{cx+d}$$