Question

If the polynomials
${x}^{2} + a {x}^{2} + 5$
and
${x}^{3} - 2 {x}^{2}$
are divided by (x+2) leave the same
remainder, find the value of a​

Answer 1

Step-by-step explanation:

Step-by-step explanation:use

Step-by-step explanation:use =4+3−12a−5

Step-by-step explanation:use =4+3−12a−5 =2−12a=R

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2)

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2)

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)Also, 3R

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)Also, 3R 1

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)Also, 3R 1

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)Also, 3R 1 +R

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)Also, 3R 1 +R 2

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)Also, 3R 1 +R 2

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)Also, 3R 1 +R 2 −28=0

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)Also, 3R 1 +R 2 −28=0∴3(2−12a)+2+4a−28=0

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)Also, 3R 1 +R 2 −28=0∴3(2−12a)+2+4a−28=0⇒6−36a+2+4a−28=0

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)Also, 3R 1 +R 2 −28=0∴3(2−12a)+2+4a−28=0⇒6−36a+2+4a−28=0⇒32a=−20

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)Also, 3R 1 +R 2 −28=0∴3(2−12a)+2+4a−28=0⇒6−36a+2+4a−28=0⇒32a=−20⇒

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)Also, 3R 1 +R 2 −28=0∴3(2−12a)+2+4a−28=0⇒6−36a+2+4a−28=0⇒32a=−20⇒ a=

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)Also, 3R 1 +R 2 −28=0∴3(2−12a)+2+4a−28=0⇒6−36a+2+4a−28=0⇒32a=−20⇒ a= 8

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)Also, 3R 1 +R 2 −28=0∴3(2−12a)+2+4a−28=0⇒6−36a+2+4a−28=0⇒32a=−20⇒ a= 8−5

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)Also, 3R 1 +R 2 −28=0∴3(2−12a)+2+4a−28=0⇒6−36a+2+4a−28=0⇒32a=−20⇒ a= 8−5

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)Also, 3R 1 +R 2 −28=0∴3(2−12a)+2+4a−28=0⇒6−36a+2+4a−28=0⇒32a=−20⇒ a= 8−5

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)Also, 3R 1 +R 2 −28=0∴3(2−12a)+2+4a−28=0⇒6−36a+2+4a−28=0⇒32a=−20⇒ a= 8−5

Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2 =16+4a−12−2 =2+4a=R 2 ⟶(2)Also, 3R 1 +R 2 −28=0∴3(2−12a)+2+4a−28=0⇒6−36a+2+4a−28=0⇒32a=−20⇒ a= 8−5

Answer 2

Answer:

f(x)=4x

3

+3x

2

−12ax−5

g(x)=2x

3

+ax

2

−6x−2

f(x) when divided by (x−1) leaves remainder as R

1

g(x) when divided by (x−2) leaves remainder as R

2

Therefore, according to Remainder theorem,

f(1)=4(1)

3

+3(1)

2

−12a(1)−5

=4+3−12a−5

=2−12a=R

1

⟶(1)

Similarly,

g(2)=2(2)

3

+a(2)

2

−6(2)−2

=16+4a−12−2

=2+4a=R

2

⟶(2)

Also, 3R

1

+R

2

−28=0

∴3(2−12a)+2+4a−28=0

⇒6−36a+2+4a−28=0

⇒32a=−20

⇒

a=

8

−5