Start by looking for integer zeros. Integer zeros are factors of the constant term.
3, - 4, and 3/2.
Practice makes perfect
We want to find all real zeros of the given polynomial function. To do so, we need to solve the equation p(x)=0.
2x^3-x^2-27x+36=0
The degree of p(x) is 3. Thus, by the Fundamental Theorem of Algebra, we know that p(x)=0 has exactly three roots. Let's find them.
Integer Roots
By the Rational Root Theorem, we know that integer roots must be factors of the constant term. Since the constant term of p(x) is 36, the possible integer roots are ± 1, ± 2, ± 3, ± 4, ± 6, ± 9, ± 12, ± 18, and ± 36. Let's check.
x
2x^3-x^2-27x+36
p(x)=2x^3-x^2-27x+36
1
2( 1)^3-( 1)^2-27( 1)+36
10 *
- 1
2( - 1)^3-( - 1)^2-27( - 1)+36
60 *
2
2( 2)^3-( 2)^2-27( 2)+36
- 6 *
- 2
2( - 2)^3-( - 2)^2-27( - 2)+36
70 *
3
2( 3)^3-( 3)^2-27( 3)+36
0 âś“
- 3
2( - 3)^3-( - 3)^2-27( - 3)+36
54 *
4
2( 4)^3-( 4)^2-27( 4)+36
40 *
- 4
2( - 4)^3-( - 4)^2-27( - 4)+36
0 âś“
6
2( 6)^3-( 6)^2-27( 6)+36
270 *
- 6
2( - 6)^3-( - 6)^2-27( - 6)+36
- 270 *
9
2( 9)^3-( 9)^2-27( 9)+36
1170 *
- 9
2( - 9)^3-( - 9)^2-27( - 9)+36
- 1260 *
12
2( 12)^3-( 12)^2-27( 12)+36
3024 *
- 12
2( - 12)^3-( - 12)^2-27( - 12)+36
- 3240 *
18
2( 18)^3-( 18)^2-27( 18)+36
10 890 *
- 18
2( - 18)^3-( - 18)^2-27( - 18)+36
- 11 466 *
36
2( 36)^3-( 36)^2-27( 36)+36
91 080 *
- 36
2( - 36)^3-( - 36)^2-27( - 36)+36
- 93 600 *
We found that 3 and - 4 are roots for p(x)=0. Therefore, (x-3) and (x+4) are factors of the polynomial. Let's use synthetic division to factor out one of these factors and thus finding the third root. Let's factor out (x-3).
rl IR-0.15cm r 3 & |rr 2 &-1 &-27 & 36
Bring down the first coefficient
rl IR-0.15cm r 3 & |rr 2 &-1 &-27 & 36 &&& & c 2 & & &
Multiply the coefficient by the divisor
rl IR-0.15cm r 3 & |rr 2 &-1 &-27 & 36 & 6 && & c 2 & & &
Using synthetic division to remove the first root, 3, left us with the following polynomial.
2x^2+5x-12
Factoring the Remaining Quadratic Factor
We will use the Quadratic Formula to find the remaining factors. To do so, we will need to identify the values of a, b, and c.
2x^2+ 5x+( - 12)=0
We can see above that a= 2, b= 5, and c= - 12. Finally, we will substitute these values into the Quadratic Formula.
We can simplify this result into two separate roots.
x=- 5± 11/4
x_1=- 5+11/4
x_2=- 5-11/4
x_1=6/4
x_2=- 16/4
x_1=3/2
x_2=- 4
We already knew that - 4 was a root, but now we have found our third root, 32. In conclusion, 3, - 4, and 32 are all real zeros of p(x)=2x^3-x^2-27x+36.
