Solve 4x^2-3x+1=a(x-1)^2+(+bx+1)

$$4 x ^ { 2 } - 3 x + 1 = a ( x - 1 ) ^ { 2 } + ( + b x + 1 )$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=-\frac{x\left(3+b-4x\right)}{\left(x-1\right)^{2}}\text{, }&x\neq 1\\a\in \mathrm{C}\text{, }&x=1\text{ and }b=1\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$4x^{2}-3x+1=a\left(x^{2}-2x+1\right)+bx+1$$

Use the distributive property to multiply a by x^{2}-2x+1.

$$4x^{2}-3x+1=ax^{2}-2ax+a+bx+1$$

Swap sides so that all variable terms are on the left hand side.

$$ax^{2}-2ax+a+bx+1=4x^{2}-3x+1$$

Subtract bx from both sides.

$$ax^{2}-2ax+a+1=4x^{2}-3x+1-bx$$

Subtract 1 from both sides.

$$ax^{2}-2ax+a=4x^{2}-3x+1-bx-1$$

Subtract 1 from 1 to get 0.

$$ax^{2}-2ax+a=4x^{2}-3x-bx$$

Combine all terms containing a.

$$\left(x^{2}-2x+1\right)a=4x^{2}-3x-bx$$

The equation is in standard form.

$$\left(x^{2}-2x+1\right)a=4x^{2}-bx-3x$$

Divide both sides by x^{2}-2x+1.

$$\frac{\left(x^{2}-2x+1\right)a}{x^{2}-2x+1}=\frac{x\left(4x-b-3\right)}{x^{2}-2x+1}$$

Dividing by x^{2}-2x+1 undoes the multiplication by x^{2}-2x+1.

$$a=\frac{x\left(4x-b-3\right)}{x^{2}-2x+1}$$

Divide x\left(-3+4x-b\right) by x^{2}-2x+1.

$$a=\frac{x\left(4x-b-3\right)}{\left(x-1\right)^{2}}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=-ax+2a+4x-\frac{a}{x}-3\text{, }&x\neq 0\\b\in \mathrm{C}\text{, }&x=0\text{ and }a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$4x^{2}-3x+1=a\left(x^{2}-2x+1\right)+bx+1$$

Use the distributive property to multiply a by x^{2}-2x+1.

$$4x^{2}-3x+1=ax^{2}-2ax+a+bx+1$$

Swap sides so that all variable terms are on the left hand side.

$$ax^{2}-2ax+a+bx+1=4x^{2}-3x+1$$

Subtract ax^{2} from both sides.

$$-2ax+a+bx+1=4x^{2}-3x+1-ax^{2}$$

Add 2ax to both sides.

$$a+bx+1=4x^{2}-3x+1-ax^{2}+2ax$$

Subtract a from both sides.

$$bx+1=4x^{2}-3x+1-ax^{2}+2ax-a$$

Subtract 1 from both sides.

$$bx=4x^{2}-3x+1-ax^{2}+2ax-a-1$$

Subtract 1 from 1 to get 0.

$$bx=4x^{2}-3x-ax^{2}+2ax-a$$

The equation is in standard form.

$$xb=-ax^{2}+4x^{2}+2ax-3x-a$$

Divide both sides by x.

$$\frac{xb}{x}=\frac{-ax^{2}+4x^{2}+2ax-3x-a}{x}$$

Dividing by x undoes the multiplication by x.

$$b=\frac{-ax^{2}+4x^{2}+2ax-3x-a}{x}$$

Divide 4x^{2}-3x-ax^{2}+2ax-a by x.

$$b=-ax+2a+4x-\frac{a}{x}-3$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{x\left(3+b-4x\right)}{\left(x-1\right)^{2}}\text{, }&x\neq 1\\a\in \mathrm{R}\text{, }&x=1\text{ and }b=1\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$4x^{2}-3x+1=a\left(x^{2}-2x+1\right)+bx+1$$

Use the distributive property to multiply a by x^{2}-2x+1.

$$4x^{2}-3x+1=ax^{2}-2ax+a+bx+1$$

Swap sides so that all variable terms are on the left hand side.

$$ax^{2}-2ax+a+bx+1=4x^{2}-3x+1$$

Subtract bx from both sides.

$$ax^{2}-2ax+a+1=4x^{2}-3x+1-bx$$

Subtract 1 from both sides.

$$ax^{2}-2ax+a=4x^{2}-3x+1-bx-1$$

Subtract 1 from 1 to get 0.

$$ax^{2}-2ax+a=4x^{2}-3x-bx$$

Combine all terms containing a.

$$\left(x^{2}-2x+1\right)a=4x^{2}-3x-bx$$

The equation is in standard form.

$$\left(x^{2}-2x+1\right)a=4x^{2}-bx-3x$$

Divide both sides by x^{2}-2x+1.

$$\frac{\left(x^{2}-2x+1\right)a}{x^{2}-2x+1}=\frac{x\left(4x-b-3\right)}{x^{2}-2x+1}$$

Dividing by x^{2}-2x+1 undoes the multiplication by x^{2}-2x+1.

$$a=\frac{x\left(4x-b-3\right)}{x^{2}-2x+1}$$

Divide x\left(-3+4x-b\right) by x^{2}-2x+1.

$$a=\frac{x\left(4x-b-3\right)}{\left(x-1\right)^{2}}$$

Solve for b

$$\left\{\begin{matrix}b=-ax+2a+4x-\frac{a}{x}-3\text{, }&x\neq 0\\b\in \mathrm{R}\text{, }&x=0\text{ and }a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$4x^{2}-3x+1=a\left(x^{2}-2x+1\right)+bx+1$$

Use the distributive property to multiply a by x^{2}-2x+1.

$$4x^{2}-3x+1=ax^{2}-2ax+a+bx+1$$

Swap sides so that all variable terms are on the left hand side.

$$ax^{2}-2ax+a+bx+1=4x^{2}-3x+1$$

Subtract ax^{2} from both sides.

$$-2ax+a+bx+1=4x^{2}-3x+1-ax^{2}$$

Add 2ax to both sides.

$$a+bx+1=4x^{2}-3x+1-ax^{2}+2ax$$

Subtract a from both sides.

$$bx+1=4x^{2}-3x+1-ax^{2}+2ax-a$$

Subtract 1 from both sides.

$$bx=4x^{2}-3x+1-ax^{2}+2ax-a-1$$

Subtract 1 from 1 to get 0.

$$bx=4x^{2}-3x-ax^{2}+2ax-a$$

The equation is in standard form.

$$xb=-ax^{2}+4x^{2}+2ax-3x-a$$

Divide both sides by x.

$$\frac{xb}{x}=\frac{-ax^{2}+4x^{2}+2ax-3x-a}{x}$$

Dividing by x undoes the multiplication by x.

$$b=\frac{-ax^{2}+4x^{2}+2ax-3x-a}{x}$$

Divide 4x^{2}-3x-ax^{2}+2ax-a by x.

$$b=-ax+2a+4x-\frac{a}{x}-3$$