Question

Divide the first polynomial by the second polynomial in each of the following, Also, write the quotient and remainder.

(i) $3x^2+4x+5,x-2$

(ii) $10x^2-7x+8,5x-3$

(iii) $5y^2-6y^2+6y-1,5y-1$

(iv) $x^4-x^3+5x,x-1$

(v) $y^4+y^2,y^2-2$

Answer 1

(i) $3x^2+4x+5,x-2$

$=3x(x-2)+10x+5$

$=3x(x-2)+10(x-2)+25$

$\therefore$ Quotient = 25

(ii) $10x^2-7x+8,5x-3$

$10x^2-6x-x+8$ $8-\frac{3}{5}$

$=2x(5x-3)-x+8$ $\frac{40-3}{5}=\frac{37}{5}$

$=2x(5x-3)-\frac{1}{5}(5x-3)+\frac{37}{5}$

$\therefore$ Quotient $=2x-\frac{1}{5}$

Remainder $=\frac{37}{5}$

(iii) $5y^3-6y^2+6y-1,5y-1$

$=y^2(5y-1)-5y^2+6y-1$

$=y^2(5y-1)-y(5y-1)+5y-1$

$=y^2(5y-1)-y(5y-1)+1(5y-1)$

$\therefore Quotient=y^2-y+1$ and Remainder = 0

(iv) $x^4-x^3+5x,x-1$

$=x^3(x-1)+5x$

$=x^3(x-1)+5(x-1)+5$

$\therefore Quotient=x^3+5,Remainder=5$

(v) $y^4+y^2,y^2-2$

$=y^2(y^2-2)+3y^2$

$=y^2(y^2-2)+3(y^2-2)+6$

$\therefore Quotient=y^2+3$ and Remainder = 6