Find whether the first polynomial is a factor of the second.
Question:

Find whether the first polynomial is a factor of the second.

(i) x + 1, 2x2 + 5x + 4

(ii) y − 2, 3y3 + 5y2 + 5y + 2

(iii) 4x2 − 5, 4x4 + 7x2 + 15

(iv) 4 − z, 3z2 − 13z + 4

(v) 2a − 3, 10a2 − 9a − 5

(vi) 4y + 1, 8y2 − 2y + 1

Solution:

(i) $\frac{2 x^{2}+5 x+4}{x+1}$

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

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

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

$\because$ Remainder $=1$

Therefore, $(\mathrm{x}+1)$ is not a factor of $2 \mathrm{x}^{2}+5 \mathrm{x}+4$

(ii) $\frac{3 y^{3}+5 y^{2}+5 y+2}{y-2}$

$=\frac{3 \mathrm{y}^{2}(\mathrm{y}-2)+11 \mathrm{y}(\mathrm{y}-2)+27(\mathrm{y}-2)+56}{\mathrm{y}-2}$

$=\frac{(\mathrm{y}-2)\left(3 \mathrm{y}^{2}+11 \mathrm{y}+27\right)+56}{\mathrm{y}-2}$

$=\left(3 \mathrm{y}^{2}+11 \mathrm{y}+27\right)+\frac{56}{\mathrm{y}-2}$

$\because$ Remainder $=56

$\therefore(\mathrm{y}-2)$ is not $a$ factor of $3 \mathrm{y}^{3}+5 \mathrm{y}^{2}+5 \mathrm{y}+2$.

$($ iii $) \frac{4 x^{4}+{ }^{2}+15}{4 x^{2}-5}$

$=\frac{x^{2}\left(4 x^{2}-5\right)+3\left(4 x^{2}-5\right)+30}{4 x^{2}-5}$

$=\frac{\left(4 x^{2}-5\right)\left(x^{2}+3\right)+30}{4 x^{2}-5}$

$=\left(x^{2}+3\right)+\frac{30}{4 x^{2}-5}$

$\because$ Remainder $=30$

Therefore, $\left(4 \mathrm{x}^{2}-5\right)$ is not a factor of $4 \mathrm{x}^{4}+7 \mathrm{x}^{2}+15$

(iv) $\frac{3 z^{2}-13 z+4}{4-z}$

$=\frac{3 z^{2}-12 z-z+4}{4-z}$

$=\frac{3 z(z-4)-1(z-4)}{4-z}$

$=\frac{(\mathrm{z}-4)(3 \mathrm{z}-1)}{4 \mathrm{z}}$

$=\frac{(4-\mathrm{z})(1-3 \mathrm{z})}{4-\mathrm{z}}$

$=1-3 \mathrm{z}$

$\because$ Remainder $=0$

$\therefore(4-\mathrm{z})$ is $a$ factor of $3 \mathrm{z}^{2}-13 \mathrm{z}+4$.

$(\mathrm{V}) \frac{10 \mathrm{a}^{2}-9 \mathrm{a}-5}{2 \mathrm{a}-3}$

$=\frac{5 \mathrm{a}(2 \mathrm{a}-3)+3(2 \mathrm{a}-3)+4}{2 \mathrm{a}-3}$

$=\frac{(2 \mathrm{a}-3)(5 \mathrm{a}+3)+4}{2 \mathrm{a}-3}$

$=(5 \mathrm{a}+3)+\frac{4}{2 \mathrm{a}-3}$

$\because$ Remainder $=4$

$\therefore(2 a-3)$ is not $a$ factor of $10 a^{2}-9 a-5$.

(vi) $\frac{8 \mathrm{y}^{2}-2 \mathrm{y}+1}{4 \mathrm{y}+1}$

$=\frac{2 \mathrm{y}(4 \mathrm{y}+1)-1(4 \mathrm{y}+1)+2}{4 \mathrm{y}+1}$

$=\frac{(4 y+1)(2 y-1)+2}{4 y+1}$

$=(2 y-1)+\frac{2}{4 y+1}$

$\because$ Remainder $=2$

$\therefore(4 y+1)$ is not $a$ factor of $8 y^{2}-2 y+1$.

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