Can you please help me factor these expressions?
Completely factor the following expressions.
Please show work.
4x^2 - 8x - 12 + 6x
144 - 9p^2
5c^2 - 24cd - 5d^2
w^2 - 17w + 42
256z^2 - 4 - 192z^2 + 3
2a^2 c^3 - 14bc^3 + 32c^3 d^2
35g^2 + 6g - 9
3j^3 - 51j^2 + 210j
Thank you so much for your help!
Ok... I tried to figure these out. This is what I got. Will you check my work?
1).
4x^2 - 8x - 12 + 6x
= 4x^2 - 2x -12
= 2(2x + 3)(x - 2)
2).
144 - 9p^2
= 9(16 - p^2)
= -9(p^2 - 4^2)
= -9(p + 4)(p - 4)
3).
5c^2 - 24cd - 5d^2
= (5c + d)(c - 5d)
4).
w^2 - 17w + 42
= (w - 3)(w + 14)
5).
256z^2 - 4 - 192z^2 + 3
= 64z^2 - 1
= (8z + 1)(8z - 1)
6).
2a^2 c^3 - 14bc^3 + 32c^3 d^2
= 2c^3(a^2 - 7b + 16d^2)
7).
35g^2 + 6g - 9
= (7g - 3)(5g + 3)
8).
3j^3 - 51j^2 + 210j
= 3j(j - 7)(j - 10)
Sure! I'd be happy to help you factor these expressions. Here are the step-by-step explanations for each expression:
1. 4x^2 - 8x - 12 + 6x:
To factor this expression, we can first combine like terms. Group the x terms together and the constant terms together:
(4x^2 + 6x) + (-8x -12)
Now, factor out the greatest common factor from each grouping:
2x(2x + 3) - 4(2x + 3)
Notice that we have a common binomial factor, (2x + 3). Factor it out:
(2x - 4)(2x + 3)
So, the completely factored form of the expression is (2x - 4)(2x + 3).
2. 144 - 9p^2:
This expression is a difference of squares since 144 is a perfect square (12^2) and 9p^2 is also a perfect square (3p)^2. Therefore, we can use the formula a^2 - b^2 = (a + b)(a - b):
(12)^2 - (3p)^2
(12 + 3p)(12 - 3p)
So, the completely factored form of the expression is (12 + 3p)(12 - 3p).
3. 5c^2 - 24cd - 5d^2:
To factor this expression, we first look for common factors among the terms. In this case, there are none. So, we need to use the method of factoring trinomials. We need to find two numbers such that their product is the product of the coefficients of the first and last terms (5 and -5d^2) and their sum is the coefficient of the middle term (-24cd). The numbers are -5 and 1. Now we can rewrite the expression as:
5c^2 - 10cd - 14cd - 5d^2
Now, factor by grouping:
(5c^2 - 10cd) - (14cd + 5d^2)
Factor out the greatest common factor from each grouping:
5c(c - 2d) - 5d(14c + d)
Now factor out the (-1) from the second grouping:
5c(c - 2d) - 5d(14c + d)
Finally, factor out the common factor of (c - 2d):
(c - 2d)(5c - 5d)
So, the completely factored form of the expression is (c - 2d)(5c - 5d).
4. w^2 - 17w + 42:
To factor this expression, we are looking for two numbers whose product is the product of the coefficients of the first and last terms (1 and 42) and whose sum is the coefficient of the middle term (-17). The numbers are -2 and -15. Now we can rewrite the expression as:
w^2 - 2w - 15w + 42
Factor by grouping:
(w^2 - 2w) - (15w - 42)
Factor out the greatest common factor from each grouping:
w(w - 2) - 3(5w - 14)
Finally, factor out the common factor of (w - 2):
(w - 2)(w - 21)
So, the completely factored form of the expression is (w - 2)(w - 21).
5. 256z^2 - 4 - 192z^2 + 3:
To factor this expression, we first combine like terms:
(256z^2 - 192z^2) + (-4 + 3)
Now simplify:
64z^2 - 1
This expression cannot be factored any further since there are no common factors among the terms. So, the factored form of the expression is 64z^2 - 1.
6. 2a^2 c^3 - 14bc^3 + 32c^3 d^2:
To factor this expression, we can first factor out the greatest common factor, which is c^3. So, rewrite the expression as:
c^3(2a^2 - 14b + 32d^2)
Now, let's factor the trinomial inside the parentheses:
2a^2 - 14b + 32d^2
We need to find two numbers whose product is the product of the coefficients of the first and last terms (2 and 32d^2) and whose sum is the coefficient of the middle term (-14b). The numbers are -2 and -16d^2. Now, rewrite the expression as:
2a^2 - 2b - 16d^2b + 32d^2
Now, factor by grouping:
(2a^2 - 2b) - 16d^2(b - 2)
Factor out the greatest common factor from each grouping:
2a^2(a - b) - 16d^2(b - 2)
Finally, factor out the common factor of (b - 2):
2a^2(a - b) - 16d^2(b - 2)
So, the completely factored form of the expression is c^3(2a^2 - 2b)(a - b) - 16d^2(b - 2).
7. 35g^2 + 6g - 9:
To factor this expression, we need to find two numbers such that their product is the product of the coefficients of the first and last terms (35 and -9) and their sum is the coefficient of the middle term (6). The numbers are 9 and -1. Now we can rewrite the expression as:
35g^2 + 9g - 10g - 9
Now, factor by grouping:
(35g^2 + 9g) - (10g + 9)
Factor out the greatest common factor from each grouping:
g(35g + 9) - 1(10g + 9)
Finally, factor out the common factor of (35g + 9):
(35g + 9)(g - 1)
So, the completely factored form of the expression is (35g + 9)(g - 1).
8. 3j^3 - 51j^2 + 210j:
To factor this expression, first, we look for a common factor among the terms. In this case, there is a common factor of j. Now, we can factor it out:
j(3j^2 - 51j + 210)
Now we can factor the trinomial inside the parentheses. We need to find two numbers such that their product is the product of the coefficients of the first and last terms (3 and 210) and their sum is the coefficient of the middle term (-51). The numbers are -6 and -35. Now we can rewrite the expression as:
j(3j^2 - 6j - 35j + 210)
Now, factor by grouping:
(j(3j^2 - 6j)) - (j(35j - 210))
Factor out the greatest common factor from each grouping:
j(3j(j - 2)) - j(35(j - 6))
Finally, factor out the common factor of (j - 2):
j(j - 2)(3j - 35)
So, the completely factored form of the expression is j(j - 2)(3j - 35).
I hope this helps! Let me know if you have any further questions.
4x^2 - 8x - 12 + 6x , looks like grouping works nicely
= 4x^2- 8x+ 6x - 12
= 4x(x-2) + 6(x-2)
= (x-2)(4x+6)
= 2(x-2)(2x+3)
144 - 9p^2 , should have recognized difference of squares
= (12-3p)(12+3kp)
5c^2 - 24cd - 5d^2
if it factors it must start with (5c ...)(c...)
and end with (... 5c)(.... c)
should have been obviously
(5c + 1)(c - 5c)
w^2 - 17w + 42
really ? what 2 numbers multiply to get +42 and have a sum of -17
They must obviously be both negative.
how about -3 and -14 ?
form the factors.
256z^2 - 4 - 192z^2 + 3 , wants to be simplified first
= 64z^2 - 1
you do it !
2a^2 c^3 - 14bc^3 + 32c^3 d^2
common factor first ...
= 2c^3( a^2 - 7b + 16d^2)
nothing else to do here
3j^3 - 51j^2 + 210j
again, common factor first
= 3j(j^2 - 17j + 70)
mmmh,, can you think of 2 numbers which when multiplied give you 70 , and when added will give you -17 ?
let me know what you get for this one
Yeahhh!
you got it