Question

Find, whether or not the first polynomial is a factor of the second :

(i) $x+1,2x^2+5x+4$

(ii) $y-2,3y^3+5y^2+5y+2$

(iii) $4x^2-5,4x^4+7x^2+15$

(iv) $4-z,3z^2-13z+4$

(v) $2a-3,10a^2-9a-5$

(vi) $4y+1,8y^2-2y+1$

Answer 1

$x+1,2x^2+5x+4$

$2x^2+5x+4=2x(x+1)+3x+4$

$=2x(x+1)+3(x+1)+1$

$\because Remainder=1$

$\therefore x+1$ is not a factor of $2x^2+5x+4$

(ii) $y-2,3y^2+5y^2+5y+2$

$3y^3+5y^2+5y+2=3y^2(y-2)+11y^2+5y+2$

$=3y^2(y-2)+11y(y-2)+27y+2$

$=3y^2(y-2)+11y(y-2)+27(y-2)+56$

$\because$ Remainder = 56

$\therefore$ y - 2 is not a factor of $3y^2+5y^2+5y+2$

(iii) $4x^2-5,4x^4+7x^2+15$

$4x^4+7x^2+15=x^2(4x^2-5)+12x^2+15$

$=x^2(4x^2-5)+3(4x^2-5)+30$

$\because$ Remainder = 30

$\therefore 4x^2-5$ is not a factor of $4x^4+7x^2+15$

(iv) $4-z,3z^2-13z+4$

$3z^2-13z+4=-3z(-z+4)-z+4$

$=-3z(-z+4)+1(-z+4)$

$\because$ Remainder = 0

$\therefore$ $4-zorz+4$ is a factor of $3z^2-13z+4$

(v) $2a-3,10a^2-9a-5$

$10a^2-9a-5=5a(2a-3)+6a-5$

$=5a(2a-3)+3(2a-3)+4$

$\because$ Remainder = 4

$\therefore 2a-3$ is not a factor of $10a^2-9a-5$

(vi) $4y+1,8y^2-2y+1$

(8y^2 - 2y + 1 = 2y (4y + 1) - 4y + 1)

$=2y(4y+1)-1(4y+1)+2$

$\because$ Remainder = 2

$\therefore 4y+1$ is not a factor of $8y^2-2y+1$