Question

show that (3x+2) is a factor of (6x^3+13x^2-4) and hence factories (6x^3+13x^2-4)​

Answer 1

Given :-

$\bf{p(x) = {6x}^{3} + {13x}^{2} - 4}$

$\bf{g(x) = 3x + 2}$

Here,

$\sf = > 3x + 2 = 0 \\ = > \boxed{ \sf \: \: \: \: \: \: \: \: \: \: \: x = \frac{ - 2}{3} }$

Now,

$\sf{p \bigg( \frac{ - 2}{3} \bigg) = {6 \bigg( \frac{ - 2}{3} \bigg)}^{3} } + {13 \bigg( \frac{ - 2}{3} \bigg)}^{2} - 4 \\ \\ \sf \: \: \: \: \: \: \: = > \frac{6 \times ( - 8)}{27} + \frac{13 \times 4}{9} - 4 \\ \\ \sf \: \: \: \: \: \: \: = > \frac{ - 48}{27} + \frac{52}{9} - 4 \\ \\ \sf \: \: \: \: \: \: \: = > \frac{ - 48 + 156 - 108}{27} \\ \\ \sf \: \: \: \: \: \: \: = > \frac{ \cancel{ - 156} \cancel{+ 156}}{27} \: \: = {0}$

Since, p(-2/3) is equal to 0. The g(x) = 3x + 2 is the factor of p(x).

Thus,

$\text{Factorisation \:of \:p(x)} :$

$\large \tt \implies {6x}^{3} + {13x}^{2} - 4 \\ \tt \implies {6x}^{3} + {4x}^{2} + {9x}^{2} + 6x - 6x - 4 \\ \tt\implies {2x}^{2} (3x + 2) + 3x(3x + 2) - 2(3x + 2)\\ \tt \implies (3x + 2)( {2x}^{2} + 3x - 2 )\\ \tt\implies(3x + 2)[ {2x}^{2} + 4x - x - 2] \\ \tt \implies (3x + 2)[2x(x + 2) - 1(x + 2)]\\ \tt\implies(3x + 2)[(x + 2)(2x - 1)] \\ \implies \boxed{ \tt( \red{3x + 2})( \red{x + 2})( \red{x - 1}) \: \: }$

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Answer 2

Please mark this answer as brainliest

$2x ^{2} + 3x - 2$

This is the quotient of long division