Factor:
A.). 4(3x+1)^2 -5(3x+1) +1
B.) 5(2-3x)^2 -28(2-3x)+15
C.) 2(3a-4)^2 - (3a-4)(a+2)-6(a+2)^2
For the first two, do you notice that the binomial appears both as a square and as a first degree expression?
When you first learn to do these types, it might be a good idea to replace the binomial with a single variable.
e.g. in the first:
let 3x+1 = a , then your equation becomes
4a^2 - 5a + 1
which factors quite nicely to:
(4a+1)(a+1)
now put your value of a back in ...
= (4(3x+1) + 1)(3x+1 + 1)
= (12x + 5)(3x+2)
try the second in the same way.
for the third I would do:
let x = 3a-4 and y = a+2 to get
2x^2 - xy - 6y^2
= (x-2y)(2x+3y)
now put our replacements back in:
= (3a-4 - 2(a+2))(2(3a-4) + 3(a+2))
= (a - 8)(9a -2)
of course you could also expand each one, simplify them, and then factor them
You MUST of course get the same answer
Oh ok thanks:)
To factor the given expressions, we can follow these steps:
A.) 4(3x+1)^2 -5(3x+1) +1
Let's assign a variable, "y", to represent the expression (3x+1):
y = 3x + 1
Now, we can rewrite the expression:
4y^2 - 5y + 1
To factor this quadratic expression, we need to find two binomials that can be multiplied together to give us the original expression.
We look for two numbers, let's call them "a" and "b", such that:
a * b = (product of the coefficient of the x^2 term) * (constant term)
a + b = (coefficient of the x term)
In this case, the coefficient of the x^2 term is 4, and the constant term is 1. So we need to find two numbers whose product is 4 * 1 = 4 and whose sum is the coefficient of the x term, which is -5.
After testing different pairs of numbers, we find that the numbers -1 and -4 satisfy the conditions:
-1 * -4 = 4
-1 + (-4) = -5
So, we can rewrite the expression as:
4y^2 - 5y + 1 = 4y^2 - 1y - 4y + 1
Now, we can group the terms:
= (4y^2 - 1y) + (-4y + 1)
Notice that we can factor out common factors from each group:
= y(4y - 1) - 1(4y - 1)
Finally, we can see that (4y - 1) is a common binomial, so we can factor it out:
= (4y - 1)(y - 1)
Substituting back y = 3x + 1:
= (4(3x + 1) - 1)((3x + 1) - 1)
= (12x + 3 - 1)(3x + 1 - 1)
= (12x + 2)(3x)
So, the factored form of the expression 4(3x+1)^2 -5(3x+1) +1 is (12x + 2)(3x).
B.) 5(2-3x)^2 -28(2-3x)+15
Similarly, let y = (2-3x):
5y^2 - 28y + 15
We need to find two numbers (a and b) whose product is 5 * 15 = 75 and whose sum is -28.
After checking different pairs of numbers, we find that -5 and -15 satisfy the conditions:
-5 * -15 = 75
-5 + (-15) = -20
So, we can rewrite the expression as:
5y^2 - 28y + 15 = 5y^2 - 5y - 23y + 15
Grouping the terms:
= (5y^2 - 5y) + (-23y + 15)
Factoring out common factors:
= y(5y - 5) - 3(23y - 5)
Factoring further:
= y(5y - 5) - 3(23y - 5)
= (5y - 3)(y - 1)
Substituting back y = (2-3x):
= (5(2-3x) - 3)((2-3x) - 1)
= (10 - 15x - 3)(2-3x - 1)
= (-15x + 7)(-3x - 1)
So, the factored form of the expression 5(2-3x)^2 -28(2-3x)+15 is (-15x + 7)(-3x - 1).
C.) 2(3a-4)^2 - (3a-4)(a+2)-6(a+2)^2
Similarly, let y = (3a-4):
2y^2 - y(a+2) - 6(a+2)^2
We can factor out the common binomial (a+2) from the expression:
= (a+2)(2y^2 - y - 6(a+2))
Now, we look for factors of 2 * -6 = -12 and pairs that add up to -1 (coefficient of y).
After testing different combinations, we find that -4 and 3 satisfy the conditions:
-4 + 3 = -1
-4 * 3 = -12
So, we can rewrite the expression as:
(a+2)(2y^2 - 4y + 3y - 6(a+2))
= (a+2)(2y(y - 2) + 3(y - 2))
= (a+2)(2y + 3)(y - 2)
Substituting back y = (3a-4):
= (a+2)(2(3a-4) + 3)(3a-4 - 2)
= (a+2)(6a - 8 + 3)(3a - 6)
= (a+2)(6a - 5)(3a - 6)
So, the factored form of the expression 2(3a-4)^2 - (3a-4)(a+2)-6(a+2)^2 is (a+2)(6a - 5)(3a - 6).