Answer :
We want to factor the polynomial
[tex]$$
f(x)=3x^4+9x^3-45x^2-57x+90
$$[/tex]
completely into linear factors.
Step 1. Factor out the greatest common factor.
Notice that each term has a factor of 3:
[tex]$$
f(x)=3\left(x^4+3x^3-15x^2-19x+30\right).
$$[/tex]
Step 2. Find a root of the quartic polynomial.
We now need to factor
[tex]$$
P(x)=x^4+3x^3-15x^2-19x+30.
$$[/tex]
By testing possible roots (using the Rational Root Theorem), we can try [tex]$x=3$[/tex]. Substituting [tex]$x=3$[/tex], we get:
[tex]$$
3^4+3\cdot3^3-15\cdot3^2-19\cdot3+30 = 81+81-135-57+30 = 0.
$$[/tex]
Since [tex]$P(3)=0$[/tex], [tex]$x=3$[/tex] is a root and [tex]$(x-3)$[/tex] is a factor.
Step 3. Divide out the factor [tex]$(x-3)$[/tex].
Divide [tex]$P(x)$[/tex] by [tex]$(x-3)$[/tex] (by synthetic division or long division):
Using synthetic division with [tex]$x=3$[/tex] and coefficients [tex]$1, 3, -15, -19, 30$[/tex]:
• Bring down 1.
• Multiply 1 by 3 and add to 3: [tex]$1\cdot3+3=6$[/tex].
• Multiply 6 by 3 and add to [tex]$-15$[/tex]: [tex]$6\cdot3-15=18-15=3$[/tex].
• Multiply 3 by 3 and add to [tex]$-19$[/tex]: [tex]$3\cdot3-19=9-19=-10$[/tex].
• Multiply [tex]$-10$[/tex] by 3 and add to 30: [tex]$-10\cdot3+30=-30+30=0$[/tex].
The quotient polynomial is
[tex]$$
x^3+6x^2+3x-10.
$$[/tex]
Step 4. Factor the cubic polynomial.
Now, consider the cubic
[tex]$$
Q(x)=x^3+6x^2+3x-10.
$$[/tex]
Test [tex]$x=1$[/tex]:
[tex]$$
1^3+6\cdot1^2+3\cdot1-10 = 1+6+3-10=0.
$$[/tex]
Since [tex]$Q(1)=0$[/tex], [tex]$x=1$[/tex] is another root, and [tex]$(x-1)$[/tex] is a factor.
Divide [tex]$Q(x)$[/tex] by [tex]$(x-1)$[/tex] using synthetic division with coefficients [tex]$1, 6, 3, -10$[/tex]:
• Bring down 1.
• Multiply 1 by 1 and add to 6: [tex]$1+6=7$[/tex].
• Multiply 7 by 1 and add to 3: [tex]$7+3=10$[/tex].
• Multiply 10 by 1 and add to [tex]$-10$[/tex]: [tex]$10-10=0$[/tex].
The quotient is the quadratic
[tex]$$
x^2+7x+10.
$$[/tex]
Step 5. Factor the quadratic polynomial.
The quadratic factors as:
[tex]$$
x^2+7x+10 = (x+2)(x+5),
$$[/tex]
since [tex]$2+5=7$[/tex] and [tex]$2\cdot5=10$[/tex].
Step 6. Write the complete factorization.
Recall the initial factor of 3 that we factored out. Combining all factors, we get:
[tex]$$
f(x)=3(x-3)(x-1)(x+2)(x+5).
$$[/tex]
Thus, the completely factored form of the polynomial is
[tex]$$
\boxed{3(x-3)(x-1)(x+2)(x+5)}.
$$[/tex]
[tex]$$
f(x)=3x^4+9x^3-45x^2-57x+90
$$[/tex]
completely into linear factors.
Step 1. Factor out the greatest common factor.
Notice that each term has a factor of 3:
[tex]$$
f(x)=3\left(x^4+3x^3-15x^2-19x+30\right).
$$[/tex]
Step 2. Find a root of the quartic polynomial.
We now need to factor
[tex]$$
P(x)=x^4+3x^3-15x^2-19x+30.
$$[/tex]
By testing possible roots (using the Rational Root Theorem), we can try [tex]$x=3$[/tex]. Substituting [tex]$x=3$[/tex], we get:
[tex]$$
3^4+3\cdot3^3-15\cdot3^2-19\cdot3+30 = 81+81-135-57+30 = 0.
$$[/tex]
Since [tex]$P(3)=0$[/tex], [tex]$x=3$[/tex] is a root and [tex]$(x-3)$[/tex] is a factor.
Step 3. Divide out the factor [tex]$(x-3)$[/tex].
Divide [tex]$P(x)$[/tex] by [tex]$(x-3)$[/tex] (by synthetic division or long division):
Using synthetic division with [tex]$x=3$[/tex] and coefficients [tex]$1, 3, -15, -19, 30$[/tex]:
• Bring down 1.
• Multiply 1 by 3 and add to 3: [tex]$1\cdot3+3=6$[/tex].
• Multiply 6 by 3 and add to [tex]$-15$[/tex]: [tex]$6\cdot3-15=18-15=3$[/tex].
• Multiply 3 by 3 and add to [tex]$-19$[/tex]: [tex]$3\cdot3-19=9-19=-10$[/tex].
• Multiply [tex]$-10$[/tex] by 3 and add to 30: [tex]$-10\cdot3+30=-30+30=0$[/tex].
The quotient polynomial is
[tex]$$
x^3+6x^2+3x-10.
$$[/tex]
Step 4. Factor the cubic polynomial.
Now, consider the cubic
[tex]$$
Q(x)=x^3+6x^2+3x-10.
$$[/tex]
Test [tex]$x=1$[/tex]:
[tex]$$
1^3+6\cdot1^2+3\cdot1-10 = 1+6+3-10=0.
$$[/tex]
Since [tex]$Q(1)=0$[/tex], [tex]$x=1$[/tex] is another root, and [tex]$(x-1)$[/tex] is a factor.
Divide [tex]$Q(x)$[/tex] by [tex]$(x-1)$[/tex] using synthetic division with coefficients [tex]$1, 6, 3, -10$[/tex]:
• Bring down 1.
• Multiply 1 by 1 and add to 6: [tex]$1+6=7$[/tex].
• Multiply 7 by 1 and add to 3: [tex]$7+3=10$[/tex].
• Multiply 10 by 1 and add to [tex]$-10$[/tex]: [tex]$10-10=0$[/tex].
The quotient is the quadratic
[tex]$$
x^2+7x+10.
$$[/tex]
Step 5. Factor the quadratic polynomial.
The quadratic factors as:
[tex]$$
x^2+7x+10 = (x+2)(x+5),
$$[/tex]
since [tex]$2+5=7$[/tex] and [tex]$2\cdot5=10$[/tex].
Step 6. Write the complete factorization.
Recall the initial factor of 3 that we factored out. Combining all factors, we get:
[tex]$$
f(x)=3(x-3)(x-1)(x+2)(x+5).
$$[/tex]
Thus, the completely factored form of the polynomial is
[tex]$$
\boxed{3(x-3)(x-1)(x+2)(x+5)}.
$$[/tex]
