Question

if p(x) =2x3+ax2+3x-5 and q(x)=x3+x2-4x+a leaves same remainder when divided by (x-2). show that a=-13/3..

Answer 1

Given
p(x)= 2x³+ax²+3x-5
q(x)= x³+x²-4x+a
both p(x) &q(x) leaves same remainder when divided by x-2.
=> (x-2) is a factor of p(x) & q(x)
=> 2 is a zero of p(x) & q(x)
•°• p(2) = q(2)

2(2³)+a(2²)+3(2)-5= (2³)+(2²)-4(2)+a
2(8)+4a+6-5= 8+4-8+a
=> 4a+17= a+4
=> 3a=-13
=> a= -(13/3)



;)
hope it helps
Comment if you need to clear something

Answer 2

$\huge{hey}$


$\huge{ \mathfrak{here \: is \: answer}}$



$\sf{solution}$


$\bf{by \: remainder \: theorem}$


$\sf{r = p(a)}$


▶
$\bf{r = p(2)}$


$\sf{given \: - \: \large \: r \tiny{1} \large= r \tiny{2}}$


▶ p(2)=2(2)³+a(2)²+3(2)-5=(2)³+(2)²-4(2)+a

▶2(8)+a(4)+6-5=8+4-8+a

▶16+4a+1=4+a

▶17+4a=4+a

▶4a-a =4-17

▶3a=-13


$\bf{a = \frac{ - 13}{3} }$

$\large{hence \: proved}$


$\huge \bf{hope \: it \: helps}$

$\huge{ \mathbb{thanks}}$