Solve f(x)=3x^4-7x^3-19x^2+49x-14

$$f ( x ) = 3 x ^ { 4 } - 7 x ^ { 3 } - 19 x ^ { 2 } + 49 x - 14$$

Factor

$$\left(x-2\right)\left(3x-1\right)\left(x^{2}-7\right)$$

Solution Steps

To factor the expression, solve the equation where it equals to 0.

$$3x^{4}-7x^{3}-19x^{2}+49x-14=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -14 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{14}{3},±14,±\frac{7}{3},±7,±\frac{2}{3},±2,±\frac{1}{3},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{4}-7x^{3}-19x^{2}+49x-14 by x-2 to get 3x^{3}-x^{2}-21x+7. To factor the result, solve the equation where it equals to 0.

$$3x^{3}-x^{2}-21x+7=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 7 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{7}{3},±7,±\frac{1}{3},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=\frac{1}{3}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{3}-x^{2}-21x+7 by 3\left(x-\frac{1}{3}\right)=3x-1 to get x^{2}-7. To factor the result, solve the equation where it equals to 0.

$$x^{2}-7=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 0 for b, and -7 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-7\right)}}{2}$$

Do the calculations.

$$x=\frac{0±2\sqrt{7}}{2}$$

Solve the equation x^{2}-7=0 when ± is plus and when ± is minus.

$$x=-\sqrt{7} x=\sqrt{7}$$

Rewrite the factored expression using the obtained roots. Polynomial x^{2}-7 is not factored since it does not have any rational roots.

$$\left(x-2\right)\left(3x-1\right)\left(x^{2}-7\right)$$

Evaluate

$$\left(x-2\right)\left(3x-1\right)\left(x^{2}-7\right)$$