First, we can factor out 2x^2 from each term:
12x^5 + 6x^3 + 8x^2 = 2x^2(6x^3 + 3x + 4)
Next, we can try to factor the expression inside the parentheses further. One possible approach is to use grouping:
6x^3 + 3x + 4 = (6x^3 + 8x^2) + (-5x^2 + 3x + 4)
= 2x^2(3x + 4) - (5x^2 - 3x - 4)
= 2x^2(3x + 4) - (5x - 4)(x + 1)
Therefore, the fully factored form of the expression is:
12x^5 + 6x^3 + 8x^2 = 2x^2(3x + 4)(5x - 4)(x + 1)
Factor completely 12x^5 + 6×^3 + 8x^2
3 answers
wrong yet again!
(3x + 4)(5x - 4)(x + 1) = 15x^3 + 23x^2 - 8x - 16
not 6x^3 + 3x + 4
(3x + 4)(5x - 4)(x + 1) = 15x^3 + 23x^2 - 8x - 16
not 6x^3 + 3x + 4
I apologize for my mistake. Here's how to correctly factor 12x^5 + 6x^3 + 8x^2:
First, we can factor out 2x^2 from each term, which gives us:
12x^5 + 6x^3 + 8x^2 = 2x^2(6x^3 + 3x + 4)
Next, we can focus on the expression inside the parentheses, which is a cubic polynomial. One possible method to factor it is to use the Rational Root Theorem, which states that any rational root of a polynomial with integer coefficients of the form a_nx^n + a_(n-1)x^(n-1) + ... + a_1x + a_0 must have the form p/q, where p is a factor of a_0 and q is a factor of a_n. In this case, a_n = 6, a_0 = 4, and the possible rational roots are ±1, ±2, ±4, ±1/2, ±3/2. We can try these values until we find a root.
By trying out some values, we can see that x = -1 is a root of the polynomial. We can verify this by dividing (6x^3 + 3x + 4) by (x + 1) using long division or synthetic division:
2x^2 - 2x + 4
x + 1 | 6x^3 + 3x + 4
-6x^3 - 6x^2
3x^2 + 3x
-3x^2 - 3x - 4
8x + 4
Therefore, we have:
6x^3 + 3x + 4 = (x + 1)(2x^2 - 2x + 4) + 8x + 4
We can now rewrite our original expression as:
12x^5 + 6x^3 + 8x^2 = 2x^2(6x^3 + 3x + 4)
= 2x^2[(x + 1)(2x^2 - 2x + 4) + 8x + 4]
= 2x^2(x + 1)(2x^2 - 2x + 4) + 16x^3 + 8x^2
Finally, we can factor 8x^2 out of the last two terms to get:
12x^5 + 6x^3 + 8x^2 = 2x^2(x + 1)(2x^2 - 2x + 4 + 8x)
First, we can factor out 2x^2 from each term, which gives us:
12x^5 + 6x^3 + 8x^2 = 2x^2(6x^3 + 3x + 4)
Next, we can focus on the expression inside the parentheses, which is a cubic polynomial. One possible method to factor it is to use the Rational Root Theorem, which states that any rational root of a polynomial with integer coefficients of the form a_nx^n + a_(n-1)x^(n-1) + ... + a_1x + a_0 must have the form p/q, where p is a factor of a_0 and q is a factor of a_n. In this case, a_n = 6, a_0 = 4, and the possible rational roots are ±1, ±2, ±4, ±1/2, ±3/2. We can try these values until we find a root.
By trying out some values, we can see that x = -1 is a root of the polynomial. We can verify this by dividing (6x^3 + 3x + 4) by (x + 1) using long division or synthetic division:
2x^2 - 2x + 4
x + 1 | 6x^3 + 3x + 4
-6x^3 - 6x^2
3x^2 + 3x
-3x^2 - 3x - 4
8x + 4
Therefore, we have:
6x^3 + 3x + 4 = (x + 1)(2x^2 - 2x + 4) + 8x + 4
We can now rewrite our original expression as:
12x^5 + 6x^3 + 8x^2 = 2x^2(6x^3 + 3x + 4)
= 2x^2[(x + 1)(2x^2 - 2x + 4) + 8x + 4]
= 2x^2(x + 1)(2x^2 - 2x + 4) + 16x^3 + 8x^2
Finally, we can factor 8x^2 out of the last two terms to get:
12x^5 + 6x^3 + 8x^2 = 2x^2(x + 1)(2x^2 - 2x + 4 + 8x)