Use the factor theorem to find a linear factor of x3 - 5x2 + 7x − 3. Hence solve the equation x3 - 5x2 + 7x…

Question

Use the factor theorem to find a linear factor of x3 - 5x2 + 7x − 3. Hence solve the equation x3 - 5x2 + 7x − 3 = 0

ScanSolve AI Tutor · Accepted Answer

To solve Question 1.2, we will use the Factor Theorem and then factorize the polynomial.

Use the Factor Theorem to find a linear factor.
Let $f(x) = x^3 - 5x^2 + 7x - 3$ .
According to the Factor Theorem, if $(x-a)$ is a factor of $f(x)$ , then $f(a) = 0$ . We test integer factors of the constant term, -3, which are $\pm 1, \pm 3$ .

Test $x=1$ :
$f(1) = (1)^3 - 5(1)^2 + 7(1) - 3$ $f(1) = 1 - 5 + 7 - 3$ $f(1) = 8 - 8$ $f(1) = 0$
Since $f(1) = 0$ , $(x-1)$ is a linear factor of $f(x)$ .
Divide the polynomial by the linear factor.
We use synthetic division to divide $x^3 - 5x^2 + 7x - 3$ by $(x-1)$ .
1 | 1 -5 7 -3 | 1 -4 3 ---------------- 1 -4 3 0
The quotient is $x^2 - 4x + 3$ .
So, $x^3 - 5x^2 + 7x - 3 = (x-1)(x^2 - 4x + 3)$ .
Factorize the quadratic expression.
Now we need to factorize the quadratic $x^2 - 4x + 3$ .
We look for two numbers that multiply to 3 and add to -4. These numbers are -1 and -3.
So, $x^2 - 4x + 3 = (x-1)(x-3)$ .
Solve the equation.
Substitute the factored quadratic back into the polynomial equation:
$(x-1)(x-1)(x-3) = 0$ $(x-1)^2(x-3) = 0$
Set each factor to zero to find the solutions:
$x-1 = 0 \implies x = 1$ $x-3 = 0 \implies x = 3$
The solutions to the equation $x^3 - 5x^2 + 7x - 3 = 0$ are $\boxed{\mathbf{x=1 \text{ (multiplicity 2)}, x=3}}$ .

Use the factor theorem to find a linear factor of x3 - 5x2 + 7x - 3. Hence solve the equation x3 − 5x2 + 7x − 3 = 0