Question

Polynomial f x is equal to ax cube + 9 x square + 4 x minus 8 when divided by x + 3 leaves -20 as the remainder find the value of a​

Answer 1

Answer:

Since (x-2) is a factor of polynomial 2x3+ax2+bx-14, we have

2(2)3+a(2)2+b(2)-14=0

⇒ 16+4a+2b-14=0

⇒4a+2b+2=0

⇒2a+b+1=0

⇒2a+b=-1 1)

On dividing by (x-3), the polynomial 2x3+ax2+bx-14 leaves remainder 52,

⇒2(3)3+a(3)2+b(3)-14=52

⇒54+9a+3b-14=52

⇒9a+3b+40=52

⇒9a+3b=12

⇒3a+b=4 (2)

Subtracting (1) and (2), we get

a=5

substituting a =5 in (1), we get

2×5+b=-1

⇒10+b=-1

⇒b=-11

Hence , a=5 and b=-11.

Answer 2

The Remainder is same whne (x−3) divides (x3−px2+x+6) & (2x3−x2−(p+3)x−6)

∴ Using Remainder Theorem

R(3)=x3−px2+x+6

          =33−p(32)+3+6

          =27−9p+3

          =36−9p

R(3)=2x3−x2−(p+3)x−6

          =2(33)−32−(p+3)3−6

          =2×27−9−3p−9−6

          =54−24−3p

          =30−3p

Remainder are same

∴36−9p=30−3p

36−30=−3p+9p

6=6p

1=p

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