Factor and find the zeros of the relation. (18 marks) (a.) y = 6 x^2 + 5x − 4
(b.) y = 4x^2 – 64

(a.)

=(6x^2-3x)+(8x-4)

=3x(2x-1)+4(2x-1)

=(2x-1)(3x+4)

zeros are x=-4/3, x= 1/2

(b.)

=4(x^2-16)

=4(x+4)(x-4)

zeros are x= -4 and x= 4

Step-by-step explanation:

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12