Question

plz answer the above pic​

Answer 1

Answer:

1. solution

Given polynomial f(x)=5x^3 +x^2−5x−1 the factor of g(x)=x+1

If x+1 is factor then x+1=0 or x=−1

Replace x in f(x) by −1 we get

f(x)=5x^3 +x^2−5x−1 g(x)=x+1

f(−1)=5(−1)^3 +(−1)^2−5(−1)−1

⇒f(−1)=−5+1+5−1

⇒f(−1)=0

So f(−1) is zero by g(x)=x+1 then g(x)=x+1 is factor of f(x)=5x^3 +x^2−5x−1

2. solution

Given polynomial f(x)=x^3 +3x^2+3x+1 and g(x)=x+1

If x+1 is factor then f(−1)=0

Replace x in p(x) by −1 we get

f(x)=x^3 +3x ^2+3x+1

or,f(−1)=(−1)^3+3(−1)^2+3(−1)+1

or,f(−1)=−1+3−3+1

or,f(−1)=0

So g(x)=x+1 is factor of f(x)=x^3 +3x^2+3x+1

3. solution

Given polynomial f(x)=x ^3 −4x^ +x+6 the factor of g(x)=x−2

If x-2 is factor then x-2=0 or x=2

Replace x in p(x) by 2 we get

f(x)=x^3−4x^2+x+6

f(2)=(2)^3−4(2)^2 +(2)+6

⇒f(2)=8−16+2+6

⇒f(2)=0

So f(x) is zero by g(x) =x-2 then g(x)=(x-2) is factor of f(x)=x ^3−4x^2 −x+6

4. solution

Given f(x)=3x^3 +x^2−20x+12 the g(x)=3x−2 is the factor

If g(x)=3x−2 is factor then g(x)=3x−2=0 or x=2/3

Replace x by 2/3 in f(x) we get

f(x)=3x^3+x^2 −20x+12

f(2/3 )=3( 2/3)^3+(2/3)^2 −20(2/3)+12

⇒f(2/3 )=3×(8×27) + (4/9)−20×2/3 +12

f(2/3 )=(8/9)+(4/9)- (40/3)+12

f(2/3)= (8+4−120+108)/9= (120−120/9) =0

So f(x)is zero by g(x) then g(x) =3x-2 is factor of f(x)=3x^3+x^2−20x+12

5 solution

Given f(x)=4x^3+20x^2+33x+18 and g(x)=2x+3

Put 2x+3=0

⟹x=−3/2

2x+3 is a factor of f(x) if f(−3/2)=0

replace x by − 3/2 in f(x) we get,

f(−3/2)=4(−3/2)^3+20(− 3/2)^2+33(−3/2)+18

⟹f(−3/2)= −4×(27/8)+20×(9/4)-(99/2)+18

⟹f(−3/2)= − (-54/4)+(180/4)-(99/2)+18

⟹f(−3/2)= (−54+180−198+72) /(4)

⟹f(− )=0

Therefore, g(x) is the factor of f(x)=4x^3+20x ^2+33x+18