Question

ਦੀ ਲੋੜ ਹੀ I — I — Alls – PRAAT

Answer 1

A

Explanation:

`{ \tt{ {30x}^(3) - {13x}^(2) - 40x - 12}}`

- For a value to be a factor of a function, the function range must be zero if the factor is substituted into it.

Testing with 5x+2

`{ \tt{5x + 2 = 0}} \\ { \tt{x = - (2)/(5) }}`

Therefore;

`{ \tt{ = 30( - (2)/(5)) {}^(3) - 13( - (2)/(5) ) {}^(2) - 40( - (2)/(5)) - 12 }} \\ \\ = { \tt{0}}`

hence 5x + 2 is a factor

Answer 2

The given polynomial is:

`30x^3-13x^2-40x-12`

Substitute x = -2/5 into the expression:

`30\left(-(2)/(5)\right)^3-13\left(-(2)/(5)\right)^2-40\left(-(2)/(5)\right)-12=0`

Therefore by the remainder theorem x = -2/5 is a root.

`\begin{gathered} x=-(2)/(5) \\ multiply\text{ both sides by 5} \\ 5x=-2 \\ 5x+2=0 \end{gathered}`

Hence, the factor is

5x + 2

A