Question

using factor theorem show that g(x) ia a factor of p(x), when - p(x)=69+11x-x²+x³ , g(x)=x+3

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Answer 1

Answer:

is the answer for using factor theorem show that g(x) ia a factor of p(x), when - p(x)=69+11x-x²+x³ , g(x)=x+3

Answer 2

Answer:

Given ⤵

➡p(x)=69+11x-x²+x³ , g(x)=x+3

To find ⤵

➡Using factor theorem to showing that g(x) is a factor of p(x).

Solution⤵

⚫If f(x) is a polynomial of degree n ≥ 1 and 'a' is any real number, then, (x-a) is a factor of f(x), if f(a)=0. Also, we can say, if (x-a) is a factor of polynomial f(x), then f(a) = 0.

➡P(x) = 69+11x-x²+x²

⚫If g(x) = x+3 is factor of p(x) then p(-3) =0

➡p(-3) = 69+11×(-3) -(-3) ²+(-3) ³

➡p(-3) = 69-33-(9) +(-27)

➡p(-3) =69-33-9-27

➡p(-3) =69-42-27

➡p(-3) =69-69

➡p(-3) =0

⚫Here according the factor theorem p(-3) is equal to zero.

✅Hence, g(x) =x+3 is factor of p(x) =69+11x-x²+x³.

By long divisione methode ⤵

⚫We can write p(x) =69+11x-x²+x³ as x³-x²+11x+69.

x+3) x³-x²+11x+69(x²-4x+23

x³+3x²

- -

_______

× -4x²+11x

-4x²-12x

+ +

_________

× 23x+69

23x+69

- -

__________

× ×

⚫Here is no remainder.

✅Hence, g(x) =x+3 is factor of p(x) =69+11x-x²+x³.

✔Extra points⤵

✡According to factor theorem, if f(x) is a polynomial of degree n ≥ 1 and 'a' is any real number, then, (x-a) is a factor of f(x), if f(a)=0. Also, we can say, if (x-a) is a factor of polynomial f(x), then f(a) = 0.

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