If the polynomials and are divided by (x+2) leave the sameremainder, find the value of a

If the polynomials and are divided by (x+2) leave the sameremainde

1.	If the polynomials and are divided by (x+2) leave the sameremainder, find the value of a
Answer» -STEP EXPLANATION:Step-by-step explanation:useStep-by-step explanation:use =4+3−12a−5Step-by-step explanation:use =4+3−12a−5 =2−12a=R Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1Step-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) 3Step-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) 2Step-by-step explanation:use =4+3−12a−5 =2−12a=R 1 ⟶(1)Similarly,g(2)=2(2) 3 +a(2) 2 −6(2)−2Step-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−12−2Step-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 2Step-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 1Step-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 2Step-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=0Step-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=0Step-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=0Step-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=−20Step-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= 8Step-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= 8−5Step-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

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If the polynomials and are divided by (x+2) leave the sameremainder, find the value of a​

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If the polynomials and are divided by (x+2) leave the sameremainder, find the value of a