Question

Using factor theorem , show that:

(i) (x + 2) is factor of x² $-$ 4.
(ii) (x $-$ 3) is factor of x³ $-$ 27.
(iii) (x $-$ 1) is factor of 2x⁴ + 9x³ + 6x² $-$ 11x $-$ 6.
(iv) (x + 4) is a factor of x² + 10x + 24.

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Let}p\left(x\right)=x^2-4 \\ \mathrm{When}x+2=0,\mathrm{we}\mathrm{have}x=-2 \\ \mathrm{Putting}x=-2\mathrm{in}p\left(x\right),\mathrm{we}\mathrm{get}: \\ p\left(-2\right)={\left(-2\right)}^2-4 \\ =4-4 \\ =0 \\ \therefore \mathrm{By}\mathrm{factor}\mathrm{theorem},\left(x+2\right)\mathrm{is}\mathrm{a}\mathrm{factor}\mathrm{of}{x}^2-4. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Let},p\left(x\right)=x^3-27 \\ \mathrm{When}x-3=0,\mathrm{we}\mathrm{have}x=3 \\ \mathrm{Putting}x=3\mathrm{in}p\left(x\right),\mathrm{we}\mathrm{get}: \\ p\left(3\right)={\left(3\right)}^3-27 \\ =27-27 \\ =0 \\ \therefore \mathrm{By}\mathrm{factor}\mathrm{theorem},\left(x-3\right)\mathrm{is}\mathrm{a}\mathrm{factor}\mathrm{of}{x}^3-27. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{Let}p\left(x\right)=2x^4+9x^3+6x^2-11x-6 \\ \mathrm{When}x-1=0,\mathrm{we}\mathrm{have}x=1 \\ \mathrm{Putting}x=3\mathrm{in}p\left(x\right),\mathrm{we}\mathrm{get}: \\ p\left(1\right)=2{\left(1\right)}^4+9{\left(1\right)}^3+6{\left(1\right)}^2-11\left(1\right)-6 \\ =2+9+6-11-6 \\ =0 \\ \therefore \mathrm{By}\mathrm{factor}\mathrm{theorem},\left(x-1\right)\mathrm{is}\mathrm{a}\mathrm{factor}\mathrm{of}2x^4+9x^3+6x^2-11x-6. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{Let},p\left(x\right)=x^2+10x+24 \\ \mathrm{When}x+4=0,\mathrm{we}\mathrm{have}x=-4 \\ \mathrm{Put}x=-4\mathrm{in}p\left(x\right)\mathrm{we}\mathrm{get}, \\ p\left(-4\right)={\left(-4\right)}^2+10\left(-4\right)+24 \\ =16-40+24 \\ =0 \\ \therefore \mathrm{By}\mathrm{factor}\mathrm{theorem},\left(x+4\right)\mathrm{is}\mathrm{a}\mathrm{factor}\mathrm{of}{x}^2+10x+24. \end{gathered}$$