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Answer :
i)\((x + 4)(x + 10)\)
\([\because (x + a)(x + b) = x^2 + (a + b)x + ab] \)
We have,
\(=(x + 4)(x + 10) \)
\( = x^2 + (4 + 10)x + (4)(10)\)
\(= x^2 + 14x + 40\)
ii) \((x + 8)(x - 10)\)
\([\because (x + a)(x + b) = x^2 + (a + b)x + ab]\)
We have,
\(= (x + 8)(x - 10) \)
\( = x^2 + (8 - 10)x + (8)(-10)\)
\(= x^2 - 2x - 80\)
iii) \((3x + 4)(3x - 5)\)
\([\because (x + a)(x + b) = x^2 + (a + b)x + ab]\)
We have,
\((3x + 4)(3x - 5) \)
\( = (3x)^2 + (4 - 5)x + (4)(5)\)
\(= 9x^2 - x + 20\)
iv) \((y^2 + \frac{3}{2} )(y^2 - \frac{3}{2} )\)
\([\because (a^2 - b^2) = (a + b)(a - b)] \)
We have,
\((y^2 + \frac{3}{2} )(y^2 - \frac{3}{2} ) \)
\( = (y^2)^2 - (\frac{3}{2} )^2\)
\(= y^4 - \frac{9}{4} \)
v) \((3 - 2x)(3 + 2x)\)
\([\because (a^2 - b^2) = (a + b)(a - b)]\)
We have,
\((3 - 2x)(3 + 2x) \)
\( = (3)^2 - (2x)^2)\)
\(= 9 - 4x^2\)