Answer:
Step-by-step explanation:
<u>For 1 and 2, follow this explanation.</u>
1) As it is the coeficient of x² is 1, we need to find two numbers whose Sum S(x)=3 and the same numbers whose product is 2, P(x)= 2
S(x) = ( ) + ( ) =3
P(x) = ( ) * ( ) =2
S(x) = ( 2 ) + ( 1 ) =3
P(x) = ( 2 ) * ( 1 ) =2
These numbers in bold will be coeficients of x.
Rewrite the Trinomial x²+ 3x +2 as x² +<u>2</u><u>x+</u><u>1</u><u>x</u>+2
So x²+3x+2 =x²+2x+1x+2
x² +2x+x+2 Put the x in evidence
x(x+2)+1(x+2) Find the Maximum Common Divider
(x+1)(x+2)
2) x²+x-6=0
Applying the same procedure. Which two integer number whose sum is 1 and whose Product is -6?
S(x) = ( 3 ) + ( -2 ) =1
P(x) = ( 3 ) * ( -2 ) =-6
Rewrite
x²+x-6
x²+3x-2x-6 <em>Group them</em>
x(x+3)-2(x+3) <em>Common Term</em>
(x+3)(x-2)
3) Since the a coeficient is ≠ 1, a little adjustment must be made on our algorithm for -2x²+4x+30=0
Firstly multiply a*c, in this case -2 * 30 = -60
Which two numbers multiplied by themselves will turn out to be -60 and whose sum is -60? Since we're working with integers factoring out may be very helpful.
P(x)= ( 10 ) * ( - 6 ) = -60
S(x) = ( 10 ) + ( -6 ) = 4
Rewriting
-2x²+10x-6x+30 <em>Group them</em>
2x(x+5) -6(x+5)
(2x-6)(x+5) <em> Inserting the -1 to finally adjust</em>
<em>-</em>1(2x-6)(x+5)
4) Explanation is the same principle as a≠1
3x²+4x-4=0
P(x) = ( 6 ) * (-2 ) = -12
S(x) = ( 6 ) + ( -2 ) = 4
3x²+4x-4 = 3x²+6x-2x-4
3x²+6x -2x+4
3x(x+2)-2(x-2)
(3x-2)(x+2)
The other ones are just applications of these, above.
