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ਦੀ ਲੋੜ ਹੀ I — I — Alls – PRAAT

ਦੀ ਲੋੜ ਹੀ I — I — Alls – PRAAT-example-1
User Joshrathke
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2 Answers

19 votes
19 votes

Answer:

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

User Amklose
by
2.9k points
16 votes
16 votes

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

User Jamsesso
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2.5k points