Question

if (x^2-4) is a factor of ax^4 +bx^3 +cx^2+dx+e then d =

Answer 1

put the value of from factor into the polynomial
${x}^{2} - 4 = 0 \\ {x}^{2} = 4 \\ x = \sqrt{4} = + 2 \: \: \: and \: - 2$
put value x= 2
$a \times {2}^{4} + b \times {2}^{3} +c \times {2}^{2} + \\ d \times 2 +e$
$16a + 8b + 4c + 2d + e = 0$
for X= -2
$16a - 8b + 4c - 2d + e = 0$
on subtracting both equations we get
$16b + 4d = 0 \\ 4d = - 16b \\ d = - 4b \: \: \: \: \: \: answer$

Answer 2

Since x2 - 1 = (x - 1) is a factor of

p(x) = ax4 + bx3 + cx2 + dx + e

∴ p(x) is divisible by (x+1) and (x-1) separately

⇒ p(1) = 0 and p(-1) = 0

p(1) = a(1)4 + b(1)3 + c(1)2 + d(1) + e = 0

⇒ a + b + c + d + e = 0 ---- (i)

Similarly, p(-1) = a (-1)4 + b (-1)3 + c (-1)2 + d (-1) + e = 0

⇒ a - b + c - d + e = 0

⇒ a + c + e = b + d ---- (ii)

Putting the value of a + c + e in eqn , we get

a + b + c + d + e = 0

⇒ a + c + e + b + d = 0

⇒ b + d + b + d = 0

⇒ 2(b+d) = 0

⇒ b + d = 0 ---- (iii)

comparing equations (ii) and (iii) , we get

a + c + e = b + d = 0

$\huge\red{brainliest\:plz}$