Asked by ANONYMUS
Directions: Find the prime factors of the polynomials
1.
a. 2a2 - 2b2
b. 6x2 - 6y2
c. 4x2 - 4
d. ax2 - ay2
e. cm2 - cn2
2.
f. st2 - s
g. 2x2 - 18
h. 2x2 - 32
i. 3x2 - 27y2
j. 18m2 - 8
3.
k. 12a2 - 27b2
l. 63c2 - 7
m. x3 - 4x
n. y3 - 25y
o. z3 - z
4.
p. 4c3 - 49c
q. 9db2 - d
r. 4a3 - ab2
s. 4a2 - 36
t. x4 - 1
5.
u. 3x2+ 6x
v. 4r2 - 4r - 48
w. x3 - 7x2 + 10x
x. 4x2 -6 x - 48
y. 16x2 - x2 v 4
Please help I don't understand how to do this pleasssseeeeee help I need to hand it in by tomorrow
1.
a. 2a2 - 2b2
b. 6x2 - 6y2
c. 4x2 - 4
d. ax2 - ay2
e. cm2 - cn2
2.
f. st2 - s
g. 2x2 - 18
h. 2x2 - 32
i. 3x2 - 27y2
j. 18m2 - 8
3.
k. 12a2 - 27b2
l. 63c2 - 7
m. x3 - 4x
n. y3 - 25y
o. z3 - z
4.
p. 4c3 - 49c
q. 9db2 - d
r. 4a3 - ab2
s. 4a2 - 36
t. x4 - 1
5.
u. 3x2+ 6x
v. 4r2 - 4r - 48
w. x3 - 7x2 + 10x
x. 4x2 -6 x - 48
y. 16x2 - x2 v 4
Please help I don't understand how to do this pleasssseeeeee help I need to hand it in by tomorrow
Answers
Answered by
Steve
Usually people don't talk about "prime factors" of polynomials, but a complete factorization should do the job. Here are a few:
a. 2a^2 - 2b^2
= 2(a^2-b^2)
= 2(a-b)(a+b)
b. 6x^2 - 6y^2
= 6(x^2-y^2)
= 2*3(x-y)(x+y)
c. 4x^2 - 4
= 4(x^2-1)
= 2*2(x-1)(x+1)
d. ax^2 - ay^2
= a(x^2-y^2)
= a(x-y)(x+y)
e. cm^2 - cn^2
= c(m^2-n^2)
= c(m-n)(m+n)
w. x^3 - 7x^2 + 10x
= x(x^2-7x+10)
= x(x-2)(x-5)
Now you try some. It looks like this is mainly a set of exercises on the difference of squares.
a. 2a^2 - 2b^2
= 2(a^2-b^2)
= 2(a-b)(a+b)
b. 6x^2 - 6y^2
= 6(x^2-y^2)
= 2*3(x-y)(x+y)
c. 4x^2 - 4
= 4(x^2-1)
= 2*2(x-1)(x+1)
d. ax^2 - ay^2
= a(x^2-y^2)
= a(x-y)(x+y)
e. cm^2 - cn^2
= c(m^2-n^2)
= c(m-n)(m+n)
w. x^3 - 7x^2 + 10x
= x(x^2-7x+10)
= x(x-2)(x-5)
Now you try some. It looks like this is mainly a set of exercises on the difference of squares.
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