can anyone help me with these:
3x2 + 17x + 10
5y8 - 125
a2 - 2ab - 15b2
4c2 - 12c + 9
5 y^8 - 125 = 5(y^8 - 25)
y^8 - 25 can be written
(y^4 - 5)(y^4 + 5)
For the others, you need to try a few factors until you find a pair that works. For example,
a^2 - 2ab - 15b^2 = (a+3b)(a-5b)
To factor the quadratic expressions 3x^2 + 17x + 10 and 4c^2 - 12c + 9, we can use a method called "factoring by grouping" or the "AC method."
Let's start with 3x^2 + 17x + 10:
Step 1: Multiply the coefficient of the x^2 term (3) with the constant term (10). The result is 30.
Step 2: We need to find two numbers that multiply together to give 30 and add up to the coefficient of the x term (17 in this case). In this case, those numbers are 5 and 6.
Step 3: Rewrite the middle term (17x) using the two numbers found in step 2.
3x^2 + 5x + 6x + 10
Step 4: Group the terms and factor by grouping:
(3x^2 + 5x) + (6x + 10)
x(3x + 5) + 2(3x + 5)
Step 5: Notice that we now have a common binomial factor, (3x + 5). Factor it out:
(3x + 5)(x + 2)
So the factored form of 3x^2 + 17x + 10 is (3x + 5)(x + 2).
Now let's move onto 4c^2 - 12c + 9:
Step 1: Multiply the coefficient of the c^2 term (4) with the constant term (9). The result is 36.
Step 2: We need to find two numbers that multiply together to give 36 and add up to the coefficient of the c term (-12 in this case). In this case, those numbers are -6 and -6.
Step 3: Rewrite the middle term (-12c) using the two numbers found in step 2.
4c^2 - 6c - 6c + 9
Step 4: Group the terms and factor by grouping:
(4c^2 - 6c) + (-6c + 9)
2c(2c - 3) - 3(2c - 3)
Step 5: Notice that we now have a common binomial factor, (2c - 3). Factor it out:
(2c - 3)(2c - 3)
So the factored form of 4c^2 - 12c + 9 is (2c - 3)(2c - 3), which can also be written as (2c - 3)^2.
Regarding the expression 5y^8 - 125, you correctly noted that it can be factored further using the difference of squares formula.
Step 1: Rewrite 125 as 5^3.
5y^8 - 5^3
Step 2: Now apply the difference of squares:
5(y^8 - 5^2)
Step 3: Notice that we have a perfect square inside the parentheses, namely (y^4 - 5)^2.
So the factored form of 5y^8 - 125 is 5(y^4 - 5)^2.
And Finally, the factorization of a^2 - 2ab - 15b^2 can be found by trying different factor pairs of the leading coefficient (1) and the constant term (-15) until we find a pair that works.
In this case, we have (a+3b)(a-5b). To confirm this, simply expand the expression (a+3b)(a-5b) using the distributive property, and you'll find that it matches the original expression a^2 - 2ab - 15b^2.
I hope this helps you understand the process of factoring these quadratic expressions! If you have any more questions, feel free to ask.