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Answer:

  • ( c - 2 ) ( 3c + 8 )

Solution:

In the given question, we have to factorise the equation:

  • 3c² + 2c - 16

Therefore, by using the middle term splitting method:

  • This method is used in equations which are in the form of ax² + bx + c. Here, we split the middle term into two terms which on multiplying gives ac and on adding or subtracting gives bx .

➙ 3c² + 2c -16

➙ 3c² - 6c + 8c - 16

➙ ( 3c² - 6c )+ (8c - 16 )

➙ 3c( c - 2 ) + 8 ( c - 2 )

➙ ( c - 2 ) ( 3c + 8 )

Factorisation of a quadratic polynomial of the type ax² + bx + c where (a ≠ 1) .

  • To factorise ax² + bx + c, we have to find two numbers whose sum is equal to the coefficient of x and product is equal to the coefficient of x² and constant term.

Consider the factorisation of 3c² + 2c – 16 .

We have to find two numbers whose sum is +2 and product (3 × 16) = 48 .

Obviously, the two numbers are 6 and 8 .

[tex]{ \qquad \sf { \dashrightarrow{ 3c {}^{2} + 2c - 16 }}}[/tex]

So, We can write it as,

[tex]{ \qquad \sf { \dashrightarrow{ 3c {}^{2} + 8c - 6c - 16 }}}[/tex]

[tex]{ \qquad \sf { \dashrightarrow{ c (3c + 8) -2 (3c + 8) }}}[/tex]

[tex]{ \qquad \sf { \dashrightarrow{ ( c -2 )(3c + 8) }}}[/tex]

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