Question

When the polynomial x3

+ 2x2

– 5ax – 7 is divided by (x + 1) it leaves the ramainder R1

, when the

polynomial x3

+ ax2

– 12x + 6 is divided by x – 2, it leave the remainder R2

if R1

– R2

= 20, then value

of ‘a’ is​

Answer 1

Answer:

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)²

⇒R2 =(2)³+a×2² −12×2+6

⇒R2=8+4a−24+6

⇒R2=4a−10

Substituting the values of R1 and R2(1) in R2+RR =6, we get

⇒2(5a−6)+(4a−10)=6

⇒10a−12+4a−10=6

⇒14a−22=6

⇒14a−28=0

⇒a=2

Step-by-step explanation:

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