Question

factorise the following 15p^4+3p^2-18​

Answer 1

Answer:

$15 {p}^{4} + 3 {p}^{2} - 18 = 0 \\ 3(5 {p}^{4} + {p}^{2} - 6) = 0 \\ 5 {p}^{4} + {p}^{2} - 6 = 0 \\ 5 {p}^{4} + 6 {p}^{2} - 5 {p}^{2} - 6 = 0 \\ {p}^{2} (5 {p}^{2} + 6) - 1(5 {p}^{2} + 6) = 0 \\ ( {p}^{2} - 1)(5 {p}^{2} + 6) = 0 \\ p = 1$

please marks as a brainlieat answer

Step-by-step explanation:

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Answer 2

$\huge{\orange{\underline{\red{\mathbf{ANSWER}}}}}$

$\implies$$\red{15p^4+3p^2-18}$

$\implies$$\red{3(5p^4+p^2-6)}$

$\implies$$\red{3(5p^4+6p^2-5p^2-6)}$

$\implies$$\red{3(p^2×(5p^2+6)-5p^2-6)}$

$\implies$$\red{3(p^2×(5p^2+6)-(5p^2+6))}$

$\implies$$\red{3(5p^2+6)×(p^2-1)}$

$\pink{now\:using\:\:a^2-b^2=(a-b)(a+b)}$

$\implies$$\orange{3(5p^2+6)×(p-1)×(p+1)}$