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

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\frac{2{\mathrm{x}}^2+5\mathrm{x}+4}{\mathrm{x}+1} \\ =\frac{2\mathrm{x}(\mathrm{x}+1)+3(\mathrm{x}+1)+1}{\mathrm{x}+1} \\ =\frac{(\mathrm{x}+1)(2\mathrm{x}+3)+1}{(\mathrm{x}+1)} \\ =(2\mathrm{x}+3)+\frac{1}{\mathrm{x}+1} \\ \because \mathrm{Remainder}=1 \\ \mathrm{Therefore},(\mathrm{x}+1)\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{factor}\mathrm{of}2{\mathrm{x}}^2+5\mathrm{x}+4 \end{gathered}$$

$$\begin{gathered} (\mathrm{ii)}\frac{{3y}^3{+5y}^2+5y+2}{\mathrm{y-2}} \\ =\frac{{\mathrm{3y}}^2(y-2)+11y(y-2)+27(y-2)+56}{\mathrm{y-2}} \\ =\frac{{(y-2)(3y}^2+11y+27)+56}{\mathrm{y-2}} \\ {=(3y}^2+11y+27)+\frac{56}{\mathrm{y-2}} \\ \because \mathrm{Remainder}=56 \\ \therefore (y-2)is\mathrm{not}a\mathrm{factor}of{\mathrm{3y}}^3{+5y}^2+5y+2. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iii}\right)\frac{4{\mathrm{x}}^4{+}^2+15}{{\mathrm{4x}}^2-5} \\ =\frac{{\mathrm{x}}^2{(4x}^2{-5)+3(4x}^2-5)+30}{{\mathrm{4x}}^2-5} \\ =\frac{({\mathrm{4x}}^2{-5)(x}^2+3)+30}{{\mathrm{4x}}^2-5} \\ {=(x}^2+3)+\frac{30}{{\mathrm{4x}}^2-5} \\ \because \mathrm{Remainder}=30 \\ \mathrm{T}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e},(4{\mathrm{x}}^2-5)\mathrm{i}\mathrm{s}\mathrm{n}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{f}4{\mathrm{x}}^4+7{\mathrm{x}}^2+15 \end{gathered}$$

$$\begin{gathered} (iv)\frac{{\mathrm{3z}}^2-13z+4}{\mathrm{4-z}} \\ =\frac{{\mathrm{3z}}^2-12z-z+4}{\mathrm{4-z}} \\ =\frac{\mathrm{3z(z-4)-1(z-4)}}{\mathrm{4-z}} \\ =\frac{(z-4)(3z-1)}{\mathrm{4-z}} \\ =\frac{(\mathrm{4-z)(1-3z)}}{\mathrm{4-z}} \\ =1-3z \\ \because \mathrm{Remainder}=0 \\ \therefore (4-z)\mathrm{is}a\mathrm{factor}of{\mathrm{3z}}^2-13z+4. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{V}\right)\frac{{\mathrm{10a}}^2-9a-5}{\mathrm{2a-3}} \\ =\frac{\mathrm{5a(2a-3)+3(2a-3)+4}}{\mathrm{2a-3}} \\ =\frac{(2a-3)(5a+3)+4}{\mathrm{2a-3}} \\ =(5a+3)+\frac{4}{\mathrm{2a-3}} \\ \because \mathrm{Remainder}=4 \\ \therefore (\; 2a-3)\mathrm{is}\mathrm{not}a\mathrm{factor}\mathrm{of}{\mathrm{10a}}^2-9a-5. \end{gathered}$$

$$\begin{gathered} (vi)\frac{{\mathrm{8y}}^2-2y+1}{\mathrm{4y+1}} \\ =\frac{\mathrm{2y}(4y+1)-1(4y+1)+2}{\mathrm{4y+1}} \\ =\frac{(4y+1)(2y-1)+2}{\mathrm{4y+1}} \\ =(2y-1)+\frac{2}{\mathrm{4y+1}} \\ \because \mathrm{Remainder}=2 \\ \therefore (4y+1)\mathrm{is}\mathrm{not}a\mathrm{factor}\mathrm{of}{\mathrm{8y}}^2-2y+1. \end{gathered}$$