Can you find the value of k?

When the remainder is #25# when #2x^4-3x^2+5# is divided by #x-k#

1 Answer
Somebody N.
Mar 26, 2018

#color(blue)(k=2)color(white)(8888) # , #color(blue)(k=-2)#

#color(blue)(k=1/2+isqrt(10))color(white)(8888)# , #color(blue)(k=1/2-isqrt(10))#

Explanation:

The remainder theorem states that, If #f(x)# is divided by #(x-a)#, Then:

#f(a)=g(x)(x-a)+r#

Where #r# is the remainder

#g(x)# is the quotient

If we substitute. #x=k#

#:.#

#f(k)=g(x)(k-k)+r#

We know the remainder is 25:

#:.#

#2k^4-3k^2+5=g(x)(0)+25#

#2k^4-3k^2+5-25=0#

#2k^4-3k^2-20=0#

Let #x=k^2#

Then:

#2x^2-3x-20=0#

Factor:

#(2x+5)(x-4)=0#

#2x+5=0=>x=-5/2#

#x-4=0=>x=4#

But #x=k^2#

#:.#

#k^2=-5/2#

#k=+-sqrt(-5/2)\ \ \ #

So:

#color(blue)(k=1/2+isqrt(10))#

#color(blue)(k=1/2-isqrt(10))#

#k^2=4#

#color(blue)(k=+-sqrt(4)=+-2)#

So

#color(blue)(k=2#

#color(blue)(k=-2)#

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