Question

Heya....

Here's the question⬇

A polynomial in x of degree three which vanishes when x=1 and x= -2 ,and has the values 4 and 28 when x = -1 and x = 2 respectively is _______
Require good explanation


Pls don't spam

Thanks.....^_^

Answer 1

I have solved the polynomial for you.
see the attachement

P(x)=3x³+4x²-5x-2

Answer 2

Heya mate,
here is your answer,
_______________

Let the polynomial of degree 3 be p(x) = ax3 + bx2 + cx + d

Where a, b, c are coefficient of x3, x2 and x respectively and d be the constant term.

It is given that when x = 1 and x = –2, the polynomial vanished off i.e., p(1) = p(–2) = 0


=> p (1) = a + b + c + d = 0 - - - - (1)
$and \: \: p( - 2) = a { ( - 2)}^{3} + b {( - 2)}^{2} + c( - 2) + d = 0$
=> - 8a + 4b - 2c + d = 0 - - - - (2)


On equating (1) and (2), we get

a+b+c+d = -8a + 4b - 2c + d
=>9a -3b + 3c = 0
=> 3a -b + c = 0
$a = \frac{b - c}{3}$


Again, according to the given condition

p (-1) = 4 and p (2) = 28
$p( - 1) = a {( - 1)}^{3} + b {( - 1)}^{2} + c( - 1) + d = 4$
=> -a + b -c + d = 4 - - - - (3)
$and \: p(2) = a {(2)}^{3} + b {(2)}^{2} + c2 + d = 28$
=> 8a + 4b +2c +d = 28 - - - - (4)


On subtracting (3) from (4), we get

8a + 4b +2c +d - (-a + b - c + d) = 28- 4
=> 9a + 3b + 3c = 24
=> 3a + b +c = 8
$= > a = \frac{s - (b + c)}{3}$


On equating both the values of a, we get

$\frac{s - (b + c)}{3} = \frac{b - c}{3}$
=> 8-b-c = b- c
=> 8 = 2b
=> b = 4
$= > a = \frac{s - (4 + c)}{3}$
=> 3a = 4-c
$a = \frac{4 - c}{3} \: or \: c = 4 - 3a \: \: \: - - - (5)$


From (1), we have

a +b+c+d =0
=> a+4+(4-3a) +d = 0
=> -2a + d = -8
=> 2a-d = 8
=> d = 2a - 8 or 8+d/2 - - - - (6)


Again from (3), we have

-a + b - c + d = 4
-a + 4 -(4 - 3a) + 2a - 8 =4
=> 4a - 8 = 4
=> 4a = 12
=> a = 3



On putting value of a in (5), we get

c = 4 - 3 (3) = -5

Again, on putting value of a in (6), we get

d = 2 (3) -8 = 2

So, the polynomial p(x) = ax3 + bx2 + cx + d

= 3x3 + 4x2 – 5x – 2

________________________________

# nikzz

HOPE U LIKE IT

CHEERS ☺☺