Question

find the value of q,f(x)=2x³+qx²–7x–12 is divisible by g(x)=x+4,hence factorise​

Answer 1

♠️Question:

find the value of q,f(x)=2x³+qx²–7x–12 is divisible by g(x)=x+4, Hence factorise....!?

♠️Answer:

Given ,

$\large{ \sf{ \bullet{ \: f(x) = 2 {x}^{3} + q {x}^{2} - 7x - 12}}}$

And,

$\large{ \sf{ \bullet{ \: g(x) = x + 4}}}$

↠f(x) is divisible by g(x)

So, remainder = 0

We know by, Euclid Divison Lemma

$\large{ \sf{ \boxed{ \green{ p(x) = g(x).q(x) + r(x)}}}} \: \\ \large{ \sf{ \to \: where \: p(x) \: = dividend}} \\ \large{ \sf{ \to \: g(x) = divisor}} \\ \large{ \sf{ \to \: q(x) = divisor}} \\ \large{ \sf{ \to \: r(x) = remainder}}$

↠When g(x) = 0

↠ p(x)= r(x)

$\large{ \sf{ \to \: g(x) = 0}} \\ \large{ \sf{ \to \: x + 4 = 0}} \\ \large{ \sf{ \to \: x = - 4}}$

↠We have , r(x)=0

↠So, p(x)= 0

So,

$\large{ \sf{ \to \:then \: p( - 4) = 2( - 4) {}^{3} + q( { - 4}^{2} ) - 7( - 4) - 12 = 0}} \\ \large{ \sf{ \to \: 2( - 64) + 16q + 28 - 12 = 0}} \\ \large{ \sf{ \to \: - 128 + 16 + 16q = 0}} \\ \large{ \sf{ \to \: 16q - 112 = 0}} \\ \large{ \sf{ \boxed{ \red{ \to \: q = 7}}}}$

↠ Then p(x)= 2x^3+7x^2-7x-12=0

↠ We have, x+4 as a factor of 2x^3+7x^2-7x-12

Other factors are,

$\large{ \sf{ \to \: q(x) = \frac{p(x)}{g(x)} }} \\ \large{ \sf{ \to \: q(x) = \frac{2 { x }^{3} + 7 {x}^{2} - 7x - 12 }{x + 4} }} \\ \sf{ \green{ \underline{ \underline{ \dag{ \: \: by \: actual \: divison}}}}} \\ \large{ \sf{ \to \: q(x) = 2 {x}^{2} - x - 3}} \\ \large{ \sf{ \to \: q(x) = 2 {x}^{2} - 3x + 2x - 3}} \\ \sf{ \green{ \underline{ \underline{ \dag{by \: middle \: term \: factorization}}}}} \\ \large{ \sf{ \to \:q(x) = x(2x - 3) + 1(2x - 3)}} \\ \large{ \sf{ \to \: q(x) = (x + 1)(2x - 3)}} \\ \\ ↠ \large{ \sf{ other \: factors \: are}} \\ \large{ \sf{ \boxed{ \purple{(x + 1) \: and \: (2x - 3)}}}}$

♠️So final answer:

$\huge{ \boxed{ \bold{ \red{q = 7}}}} \\ \\ \huge{ \boxed{ \bold{ \red{p(x) = (x + 4)(x + 1)(2x - 3)}}}} \\ \sf{ \therefore{ \green{ \underline{ \underline{hence \: factorised}}}}}$