Q.Using synthetic division, find the equation whose roots are $2$ less than the roots of $x^3-6x^2+10x-3=0$.

Step 1. Coefficients of $f(x)=x^3-6x^2+10x-3$: $1,-6,10,-3$. Divide synthetically by $(x-2)$ ($a=2$):

$b_0=1$; $b_1=-6+2(1)=-4$; $b_2=10+2(-4)=2$; remainder $R_0=-3+2(2)=1$.

Quotient: $x^2-4x+2$.

Step 2. Divide this quotient again by $(x-2)$: coefficients $1,-4,2$: $b_0=1$; $b_1=-4+2(1)=-2$; remainder $R_1=2+2(-2)=-2$.

Quotient: $x-2$.

Step 3. Divide this quotient again by $(x-2)$: coefficients $1,-2$: $b_0=1$; remainder $R_2=-2+2(1)=0$.

Step 4. The leading coefficient stays $R_3=1$ throughout. Reading the successive remainders from the last obtained back to the first gives the coefficients of the transformed equation in $y=x-2$, constant term first:

Step 5. So the required equation is $y^3-2y+1=0$.

Step 6 (check by direct substitution). With $x=y+2$: $(y+2)^3-6(y+2)^2+10(y+2)-3$. Now $(y+2)^3=y^3+6y^2+12y+8$; $-6(y+2)^2=-6y^2-24y-24$; $10(y+2)=10y+20$. Summing: $y^3+(6-6)y^2+(12-24+10)y+(8-24+20-3)=y^3-2y+1$, confirming Step 5 exactly.

Translation of roots via Horner's process: three successive synthetic divisions by $(x-2)$, reading off the transformed coefficients from the remainders.

• Reading the remainders in the wrong order (they must be read from the last division back to the first to get the new equation's coefficients, constant term first).
• Sign confusion about whether the shift is $x=y+2$ or $x=y-2$ when 'diminishing the roots by $2$' -- diminishing means the new roots equal the old ones minus $2$, so $y=x-2$, i.e. $x=y+2$.