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Factor each polynomial if possible. If the polynomial cannot be factored using integers, write 'prime,'
1) 12q^2+34q-28
2) 18h^2+15h-18
3) 12p^2-22p-20
4) 8z^2+20z-48



Answer :

apologiabiology
first one,
12q^2+34q-28
divide whole thing by 2
6q^2+17q-14
use trial and error and get (2x+7)(3x-2)
factored out form is (2)(2x+7)(3x-2)

2.
divide by 3
6h^2+5h-6
trial and error and get (2x+3)(3x-2)
so the factored form is (3)(2x+2)(3x-2)


3. divide by 2
6p^2-11x-10
use trial and error and get (2x-5)(3x+2)
the factored out form is (2)(2x-5)(3x+2)

4.divide by 4
2z^2+5z-12
use trial and error and get (x+4)(2x-3)
the factored form is (4)(x+4)(2x-3)

to factor the basic thing is
ax^2+bx+c
b=x+y
ac=xy
solve for x and y
so exg 2z^2+5z-12
a=2
b=5
c=-12
2 times -12=-24

then factctor -24 and find factors that add up to 5
-1, 24
-2, 12
-3, 8
-4,6
now add
-1+24=23
-2+12=10
-3+8=5 check
-4+6=2
the numbers are -3 and 8
split it into such

2z^2+8z-3z-12
group (2z^2+8z)+(-3z-12)
undistribute using distributive property
ab+ac=a(b+c)
(2z)(z+4)+(-3)(z+4)
now reverse distribute again
((2z)+(-3))(z+4)
(2z-3)(z+4)

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