Determination which of the following polynomials has (x+1) as a factor (Iv) x^3-x^2-(3-√3)x+√3 friends please give me answers please I want the answer
Answers
Answer:
Apply remainder theorem
x+1=0
x=−1
Put the value of x=−1 in all equations.
(i) x
3
+x
2
+x+1=(−1)
3
+(−1)
2
+(−1)+1=−1+1−1+1=0
Then x+1 is the factor of equation
(ii) x
4
+x
3
+x
2
+x+1=(−1)
4
+(−1)
3
+(−1)
2
+(−1)+1=1−1+1−1+1=1
This is not zero.Then x+1 is not the factor of equation
(iii) x
4
+3x
3
+3x
2
+x+1=(−1)
4
+3(−1)
3
+3(−1)
2
+(−1)+1=1
This is not zero.Then x+1 is not the factor of equation
(iv)x
3
−x
2
−(2+
2
)x+
2
=(−1)
3
−(−1)
2
−(2+
2
)(−1)+
2
=−1−1+2−
2
+
2
=0
This is zero. Then x+1 is the factor of equation
Answer:
Let the given polynomial be p(x) = 4x2 – 4x – 8
To find the zeroes, take p(x) = 0
Now, factorise the equation 4x2 – 4x – 8 = 0
4x2 – 4x – 8 = 0
4(x2 – x – 2) = 0
x2 – x – 2 = 0
x2 – 2x + x – 2 = 0
x(x – 2) + 1(x – 2) = 0
(x – 2)(x + 1) = 0
x = 2, x = -1
So, the roots of 4x² – 4x – 8 are -1 and 2.
Relation between the sum of zeroes and coefficients:
-1 + 2 = 1 = -(-4)/4 i.e. (- coefficient of x/ coefficient of x²)
Relation between the product of zeroes and coefficients:
(-1) × 2 = -2 = -8/4 i.e (constant/coefficient of x²)