Question

if ax^3+bx^2+x-6 has x-2 as a factor and leaves a remainder 4 when divided by x-2 ( class 9 math ch polynomials)

Answer 1

Answer:

question is wrong x+2 is the factor of the given equation. pls. check

Step-by-step explanation:

Let p(x)= ax³+bx²+x-6

Since (x+2) is a factor of p(x), then by Factor theorem p(−2)=0

⇒a(−2)³+b(−2)²+(−2)−6=0

⇒−8a+4b−8=0

⇒−2a+b=2 ...(i)

Also when p(x) is divided by (x-2) the remainder is 4, therefore by Remainder theorem p(2)=4

⇒a(2)³+b(2)²+2−6=4

⇒8a+4b+2−6=4

⇒8a+4b=8

⇒2a+b=2 ...(ii)

Adding equation (i) and (ii), we get

(−2a+b)+(2a+b)=2+2

⇒2b=4⇒b=2

Putting b=2 in (i), we get

−2a+2=2

⇒−2a=0⇒a=0

Hence, a=0 and b=2

Answer 2

Answer:

let p(x) = ax³+bx²+x-6

a/c to question

(x+2) is the factor of p(x), and we know this is possible only when p(-2) = 0

so, p(2) = a(-2)³ + b(-2)² - 2 - 6 = 0

→ -8a + 4b - 8 = 0

→ 2a - b + 2 = 0. ________(1)

again, question said that if we p(x) is divided by (x-2) then it leaves remainder 4.

so p(2) = a(2)³ + b(2)² + 2 - 6 = 4

→ 8a + 4b - 4 = 4

2a + b - 2 = 0 _________(2)

solve equation 4a = 0

→ a = 0 and b = -2

Then, equation will be 2x² + x - 6