Question

3 Divide 2²-32²-52²16×19 by x+1 and find remainder​

Answer 1

Step-by-step explanation:

Let p(x)=x3+2x2−5ax−7

and q(x)=x3+ax2−12x+6 be the given polynomials,

Now, R1= Remainder when p(x) is divided by x+1.

⇒R1=p(−1)

⇒R1=(−1)3+2(−1)2−5a(−1)−7[∵p(x)=x2+2x2−5ax−7]

⇒R1=−1+2+5a−7

⇒R1=5a−6

And R2= Remainder when q(x) is divided by x-2

⇒R1=q(2)

⇒R2=(2)3+a×22−12×2+6[∵q(x)=x2+ax2−12x−6]

⇒R2=8+4a−24+6

⇒R2=4a−10

Substituting the values of R1 and R2 in 2R1+R2=6, we get

⇒2(5a−6)+(4a−10)=6

⇒10a−12+4a−10=6

⇒14a−22=6

⇒14a−28=0

⇒a=2

Answer 2

Answer:

When 3^2 is divided by 19, remainder = 9 ... Cyclicity of 3 = 3, 9, 7 and 1 ... 243 , when 243/19 , we get remainder as 15

Step-by-step explanation:

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