Answer :
We start with the polynomial
[tex]$$5x^3 - 19x^2 + 12x.$$[/tex]
Step 1. Factor out the common factor
Notice that each term has a factor of [tex]$x$[/tex]. Factor [tex]$x$[/tex] out:
[tex]$$5x^3 - 19x^2 + 12x = x \Bigl(5x^2 - 19x + 12\Bigr).$$[/tex]
Step 2. Factor the quadratic
Now, we need to factor the quadratic
[tex]$$5x^2 - 19x + 12.$$[/tex]
We look for two numbers that multiply to [tex]$5 \cdot 12 = 60$[/tex] and add to [tex]$-19$[/tex]. The two numbers that satisfy these conditions are [tex]$-15$[/tex] and [tex]$-4$[/tex], because
[tex]$$(-15) \cdot (-4) = 60$$[/tex]
[tex]$$(-15) + (-4) = -19.$$[/tex]
Rewrite the quadratic by decomposing the middle term:
[tex]$$5x^2 - 19x + 12 = 5x^2 - 15x - 4x + 12.$$[/tex]
Now, factor by grouping:
- Group the first two terms and the last two terms:
[tex]$$5x^2 - 15x - 4x + 12 = (5x^2 - 15x) + (-4x + 12).$$[/tex]
- Factor out the common factors in each group:
[tex]$$5x^2 - 15x = 5x(x - 3),$$[/tex]
[tex]$$-4x + 12 = -4(x - 3).$$[/tex]
- Now factor out the common binomial [tex]$(x-3)$[/tex]:
[tex]$$5x(x-3) - 4(x-3) = (x-3) \Bigl(5x-4\Bigr).$$[/tex]
Thus, the quadratic factors as
[tex]$$5x^2 - 19x + 12 = (x-3)(5x-4).$$[/tex]
Step 3. Write the fully factored form
Substituting back into the factored form with the extracted [tex]$x$[/tex], we get:
[tex]$$5x^3 - 19x^2 + 12x = x (x-3)(5x-4).$$[/tex]
Step 4. Find the roots
To find the roots of the polynomial, set each factor equal to zero:
1. From [tex]$x = 0$[/tex], we have:
[tex]$$x = 0.$$[/tex]
2. From [tex]$x-3=0$[/tex], we have:
[tex]$$x = 3.$$[/tex]
3. From [tex]$5x-4=0$[/tex], we have:
[tex]$$5x = 4 \quad \Rightarrow \quad x = \frac{4}{5}.$$[/tex]
Final Answer
The factored form of the polynomial is
[tex]$$x (x-3)(5x-4),$$[/tex]
and the roots are
[tex]$$x = 0, \quad x = 3, \quad x = \frac{4}{5}.$$[/tex]
[tex]$$5x^3 - 19x^2 + 12x.$$[/tex]
Step 1. Factor out the common factor
Notice that each term has a factor of [tex]$x$[/tex]. Factor [tex]$x$[/tex] out:
[tex]$$5x^3 - 19x^2 + 12x = x \Bigl(5x^2 - 19x + 12\Bigr).$$[/tex]
Step 2. Factor the quadratic
Now, we need to factor the quadratic
[tex]$$5x^2 - 19x + 12.$$[/tex]
We look for two numbers that multiply to [tex]$5 \cdot 12 = 60$[/tex] and add to [tex]$-19$[/tex]. The two numbers that satisfy these conditions are [tex]$-15$[/tex] and [tex]$-4$[/tex], because
[tex]$$(-15) \cdot (-4) = 60$$[/tex]
[tex]$$(-15) + (-4) = -19.$$[/tex]
Rewrite the quadratic by decomposing the middle term:
[tex]$$5x^2 - 19x + 12 = 5x^2 - 15x - 4x + 12.$$[/tex]
Now, factor by grouping:
- Group the first two terms and the last two terms:
[tex]$$5x^2 - 15x - 4x + 12 = (5x^2 - 15x) + (-4x + 12).$$[/tex]
- Factor out the common factors in each group:
[tex]$$5x^2 - 15x = 5x(x - 3),$$[/tex]
[tex]$$-4x + 12 = -4(x - 3).$$[/tex]
- Now factor out the common binomial [tex]$(x-3)$[/tex]:
[tex]$$5x(x-3) - 4(x-3) = (x-3) \Bigl(5x-4\Bigr).$$[/tex]
Thus, the quadratic factors as
[tex]$$5x^2 - 19x + 12 = (x-3)(5x-4).$$[/tex]
Step 3. Write the fully factored form
Substituting back into the factored form with the extracted [tex]$x$[/tex], we get:
[tex]$$5x^3 - 19x^2 + 12x = x (x-3)(5x-4).$$[/tex]
Step 4. Find the roots
To find the roots of the polynomial, set each factor equal to zero:
1. From [tex]$x = 0$[/tex], we have:
[tex]$$x = 0.$$[/tex]
2. From [tex]$x-3=0$[/tex], we have:
[tex]$$x = 3.$$[/tex]
3. From [tex]$5x-4=0$[/tex], we have:
[tex]$$5x = 4 \quad \Rightarrow \quad x = \frac{4}{5}.$$[/tex]
Final Answer
The factored form of the polynomial is
[tex]$$x (x-3)(5x-4),$$[/tex]
and the roots are
[tex]$$x = 0, \quad x = 3, \quad x = \frac{4}{5}.$$[/tex]
