Question

if
$x { }^{2} - 1$
is a factor of
$px {}^{4 } + qx {}^{3} + rx {}^{2} + sx + 4$
then show that
$p + r + 4 = q + s = 0$
​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Let f(x) = px⁴+qx³+rx²+sx+4

It is given that,

If (x²-1) is a factor of f(x).

(x²-1²) = (x+1)(x-1)

Therefore,

(x+1), (x-1) are factors of f(x).

i ) If (x-1) is a factor of f(x)

then f(1) = 0

=> p+q+r+s+4 = 0

=> (p+r+4)+(q+s) = 0 -----(1)

ii) If (x+1) is a factor of f(x)

then f(-1) = 0

=> p(-1)⁴+q(-1)³+r(-1)²+s(-1)+4=0

=> p-q+r-s+4=0

=> p+r+4 = q+s ----(2)

Substitute (2) in (1) , we get

=> (q+s)+(q+s) = 0

=> 2(q+s) = 0

=> q+s=0 ---(3)

now put (3) in equation (1) , we

get

p+r+4 = 0 ---(4)

Therefore,

If (x²-1) is a factor of f(x),

p+r+4 = q+s = 0 [ from (3)&(4)]

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