Question

if p (x) = to X square + bx + c is exactly divisible by (2x-1)and (X + 2 ) , p(2) =12 then a and b and c are:
a) 2,3,-2
b) 2,-2,3
c) 3,-2,2
d) 3,2,-2
Please answer fast

Answer 1

$\huge{\fbox{\fbox{\green{\mathfrak{ANSWER}}}}}$

$\huge\boxed{\underline{\mathcal{\red{(a)}\green{2,}\pink{3,}\orange{-2}\blue{.}\pink{.}}}}$

$\huge\blue{\underbrace{\overbrace{\ulcorner{\mid{\overline{\underline{EXPLANATION }}}}}}}$

Given,

${\bf{\green{=>p(x)=ax^2+bx+c... EQ1}}}$

${\bf{\red{=>p(2)=12}}}$

Sub,option (a) in EQ1

${\bf{\blue{=>p(x)=2x^2+3x-2}}}$

${\bf{\pink{=>p(2)=2(2)^2+3(2)-2}}}$

${\bf{\green{=>p(2)=8+6-2}}}$

${\bf{\orange{=>p(2)=14-2}}}$

${\bf{\blue{=>p(2)=12}}}$

Now,

${\bf{\red{divide }}}$${\bf{\green{it}}}$${\bf{\pink{by}}}$

${\bf{\blue{(2x-1)}}}$

${\bf{\red{=>(2(2)-1)}}}$

${\bf{\blue{=>(4-1)}}}$

${\bf{\pink{=>3}}}$

And

${\bf{\green{=>(x+2)}}}$

${\bf{\orange{=>(2+2)}}}$

${\bf{\blue{=>4}}}$

${\bf{\red{p(2)=12}}}$${\bf{\green{divisible}}}$${\bf{\pink{by}}}$${\bf{\blue{both}}}$ ${\bf{\green{3}}}$${\bf{\red{and}}}$${\bf{\blue{4}}}$

${\bf{\pink{so, }}}$

${\bf{\green{a=2,}}}$${\bf{\orange{b=3,}}}$${\bf{\blue{c=-2}}}$

${\bf{\red{and}}}$

${\bf{\green{X=2}}}$

Answer 2

Answer:

Hope it helps you.... Refers to the attachment.... Thanks....