I need help with these two problems please.
Completely factor the polynomial.
8x^3+16^2+8x+16
Factor the trinomial completely.
y^3-14y^2+48y
I did the first one here:
http://www.jiskha.com/display.cgi?id=1364169292
2nd:
= y(y^2 - 14y + 48)
=y(y - 6)(y - 8)
Sure! Let's start with the first problem, which involves factoring a polynomial.
To completely factor the polynomial 8x^3 + 16x^2 + 8x + 16, we can use the factoring by grouping method.
Step 1: Group the terms in pairs:
(8x^3 + 16x^2) + (8x + 16)
Step 2: Factor out the greatest common factor (GCF) from each pair:
8x^2(x + 2) + 8(x + 2)
Step 3: Notice that both terms have a common factor of (x + 2). Now, factor out the (x + 2) from both terms:
8(x^2) (x + 2) + 8(x + 2)
Step 4: Finally, factor out the common factor of 8:
8(x + 2)(x^2 + 1)
Therefore, the polynomial 8x^3 + 16x^2 + 8x + 16 can be completely factored as 8(x + 2)(x^2 + 1).
Now let's move on to the second problem, which involves factoring a trinomial.
To completely factor the trinomial y^3 - 14y^2 + 48y, we can use the factoring by grouping method.
Step 1: Factor out the greatest common factor (GCF) from all terms:
y(y^2 - 14y + 48)
Step 2: Look for two numbers whose product is equal to the constant term (48) and whose sum is equal to the coefficient of the middle term (-14). In this case, the numbers are -6 and -8.
Step 3: Rewrite the middle term (-14y) using the two numbers identified in Step 2:
y(y^2 - 6y - 8y + 48)
Step 4: Group the terms:
y((y^2 - 6y) - (8y - 48))
Step 5: Factor out the common factors from each pair of terms:
y(y(y - 6) - 8(y - 6))
Step 6: Notice that both terms now have a common factor of (y - 6). Now, factor out the (y - 6) from both terms:
y(y - 6)(y - 8)
Therefore, the trinomial y^3 - 14y^2 + 48y can be completely factored as y(y - 6)(y - 8).
I hope this helps! If you have any further questions, feel free to ask.