Question

find the HCF and LCM of (2x²+x-15) and (x⁴+27x)

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Answer 1

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$</p><p>\begin{gathered}f(x) = 40x(2 {x}^{2} - 5x - 3) \\ \: \: \: \: \: \: = 40x(2 {x}^{2} - 6x + x - 3) \\ = 40x(2x(x - 3) + 1(x - 3)) \\ = 40x(2x + 1)(x - 3) \\ = 2 \times 2 \times 2 \times 5x(2x + 1)(x - 3) \\ \\ g(x) = 35 ({x}^{4} - 27x) \\ = 35x( {x}^{3 } - 27) \\ = 35x( {x}^{3} - {3}^{3} ) \\ = 35x(x - 3)( {x}^{2} - 3x + 9) \\ = 5 \times 7 \times x(x - 3)( {x}^{2} - 3x + 9) \\ \\ so \: hcf \: is \: = 5x(x - 3) \\ lcm = {2}^{3} \times 5 \times 7 \times x \\ (2x + 1)(x - 3)( {x}^{2} - 3x + 9) \\ = 280x(2x + 1)(x - 3)( {x}^{2} - 3x + 9)\end{gathered}f(x)=40x(2x2−5x−3)=40x(2x2−6x+x−3)=40x(2x(x−3)+1(x−3))=40x(2x+1)(x−3)=2×2×2×5x(2x+1)(x−3)g(x)=35(x4−27x)=35x(x3−27)=35x(x3−33)=35x(x−3)(x2−3x+9)=5×7×x(x−3)(x2−3x+9)sohcfis=5x(x−3)lcm=23×5×7×x(2x+1)(x−3)(x2−3x+9)=280x(2x+1)(x−3)(x2−3x+9)$