Here we are provided with an polynomial of degree 3 and we have to factorise it first and then we have to compute it's zeroes.
Given Polynomial:
Taking common x from all the 3 terms,
➝ x(x² + 3x - 10)
Factorising through Middle term factorisation,
➝ x(x² + 5x - 2x - 10)
➝ x{x(x + 5) - 2(x + 5)}
➝ x(x - 2)(x + 5)
Hence, Factorised !!
Now equating to 0 to find the zeroes of the poly.
➝ x(x - 2)(x + 5) = 0
That means,
So, the zeroes of the polynomial is 0, 2 or -5
And we are done....
Carry On Learning
The table is missing in the question. The table is provided here :
Group 1 Group 2
34.86 64.14 mean
21.99 20.46 standard deviation
7 7 n
Solution :
a). The IV or independent variable = Group 1
The DV or the dependent variable = Group 2
b).
Therefore,
t = -2.579143
Now,
df = 7 - 1
= 6
Therefore the value of p :
= 0.020908803
The p value is 0.0209
So we reject the null hypothesis and conclude that