t^3-12t+16 what is the factoring of this trinomial?

Question

How to  decompose in factors the trinomial of my question?

 

Answer 1

Given trinomial t^3-12t+16

Substitute t = 2 in given trinomial.

= 2^3-12*2+16

= 8-24+16

= 0

So (t-2) is a factor.

By synthetic division method

t^2+2t-8

Now factorise t^2+2t-8

t^2+4t-2t-8

t(t+4)-2(t+4)

(t+4)(t-2)

t^2+2t-8 = (t+4)(t-2)

Factoring of t^3-12t+16 = (t-2)(t+4)(t-2)

 

Answer 2

The cubic polynomial is t³- 12t + 16 = 0

From factor theorem when f(c) = 0 then x - c is factor of polynomial.

In this case f(c) = 0 then t - c is factor of the polynomial.

Identify Rational Zeros  

Usually it is not practical to test all possible zeros of a polynomial function using only synthetic substitution. The Rational Zero Theorem can be used for finding the some possible zeros to test.

t³- 12t + 16 = 0

If p/q is a rational zero, then p is a factor of 16 and q is a factor of 1.

The possible values of p are   ± 1,   ± 2, ± 4, ± 8 and   ± 16.

The possible values for q are ± 1.

So, p/q =   ± 1,   ± 2, ± 4, ± 8 and   ± 16.

Make a table for the synthetic division and test possible  zeros.

p/q	1	0	-12	16
1	1	1	-11	5
2	1	2	-8	0

Since f(2) = 0,  t  =  2 is a zero. The depressed polynomial is   t²+ 2t - 8 = 0

Since the depressed polynomial of this zero, t²+ 2t - 8, is quadratic, use the  Factorization to find the roots of the related quadratic equation

t²+ 2t - 8 = 0

t²+ 4t -2t - 8 = 0

t(t + 4) - 2(t + 4) = 0

(t + 4)(t  - 2) = 0

t + 4 = 0 and t - 2 = 0

t = - 4 and t = 2

Zeros of the polynomial is t = -4 , 2 and 2.

Factoring of t³- 12t + 16 = (t + 4) (t - 2)(t - 2).