Question

factorise 2t^3-5t^2-19t+42​

Answer 1

hope this will help you!!!

Answer 2

Answer:

Required factors are (t-2),(t-3),(2t-7)

Step-by-step explanation:

Given polynomial is $2 {t}^{3} - 5 {t}^{2} - 19t + 42$

Let,

$f(t) = 2 {t}^{3} - 5 {t}^{2} - 19t + 42$

We want to factorise given polynomial.

We can write

$2 {t}^{3} - 5 {t}^{2} - 19t + 42 \\ = 2 {t}^{3} - (4 + 1) {t}^{2} - 19t + 42 \\ =2 {t}^{3} - 4 {t}^{2} - {t}^{2} - 19t + 42 \\ = 2 {t}^{2} (t - 2) - {t}^{2} - 19t + 42 \\ = 2 {t}^{2} (t - 2) - {t}^{2} - ( - 2 + 21)t + 42 \\ = 2 {t}^{2}(t - 2) - {t}^{2} + 2t - 21t + 42 \\ = 2 {t}^{2} (t - 2) - t(t - 2) - 21(t - 2) \\ = (t - 2)(2 {t}^{2} - t - 21) \\ = (t - 2)(2 {t}^{2} - 7t + 6t - 21) \\ = (t - 2)(t(2t - 7) + 3(2t - 7)) \\ = (t - 2)(2t - 7)(t + 3)$

All factors of 2t³-5t²-19t+42 are (t-2),(t-3),(2t-7).

Here we applied middle term method.

This is a problem Factorization of Algebra.

Some important Algebra formulas.

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

a² − b² = (a + b)(a − b)

a² + b² = (a + b)² − 2ab

a² + b² = (a − b)² + 2ab

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

Know more about Algebra,

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