3.06 Solving polynomial equations
Concept summary
A polynomial equation is an equation with polynomial expressions on both sides of the equation.
The standard form of a polynomial equation is given by p_nx^n+p_{n-1}x^{n-1}+p_{n-2}x^{n-2}+\ldots+p_2x^2+p_1x+p_0=0 where n is a positive integer and p_n,p_{n-1},p_{n-2},\ldots,p_2,p_1,p_0 are real numbers.
We can find the roots of this polynomial equation by factoring. The maximum number of roots will be n for a polynomial of degree n.
Given P\left(x\right)=p_nx^n+p_{n-1}x^{n-1}+p_{n-2}x^{n-2}+\ldots+p_2x^2+p_1x+p_0, where the coefficients are all integers. If P\left(x\right) has any rational roots then they must be of the form \dfrac{p}{q} where p is a factor of the constant term p_0 and q is a factor of the leading coefficient p_n.
Worked examples
Example 1
Solve the equation \left(4x^2-81\right)\left(x^2-3x-10\right)=0.
Approach
First, we determine whether the expression on the left-hand side of the equation can be expressed as a product of linear factors. If so, we can apply some factoring techniques.
Solution
The equation contains a factored expression on the left-hand side and 0 on the right-hand side. The factors are both of degree 2. We need to express these quadratic factors as linear factors, if possible.
Observe that 4x^2-81=(2x)^2-(9)^2. This means that 4x^2-81 is a difference of two squares. Now,
| \displaystyle a^2-b^2 | \displaystyle = | \displaystyle (a-b) (a+b) | Formula of difference of two squares |
| \displaystyle (2x)^2-(9)^2 | \displaystyle = | \displaystyle (2x-9) (2x+9) | Substitute a=2x, b=9 |
So we have 4x^2-81=(2x-9) (2x+9).
We know that x^2-3x-10 is a quadratic trinomial with a leading coefficient 1 and the constant term is -10. The factors of -10 are \pm1,\pm2,\pm5,\pm10. Observe that -5 and 2 when multiplied is -10 and when added is -3. This implies that x^2-3x-10=(x-5)(x+2).
The equation \left(4x^2-81\right)\left(x^2-3x-10\right)=0 becomes (2x-9) (2x+9)(x-5)(x+2)=0
The Zero Product Property states that if A \cdot B \cdot C \cdot D=0, then A =0, B=0, C=0 or D=0.
So we get the following equations:2x-9=0, 2x+9=0,x-5=0 \text{ or }x+2=0 Solving each equation for x, we obtain the following solutions: x=\dfrac{9}{2}, x=-\dfrac{9}{2},x=5,x=-2
Reflection
We can check the answer by substituting each solution to \left(4x^2-81\right)\left(x^2-3x-10\right)=0.
- \left(4\left(\dfrac{9}{2}\right)^2-81\right)\left(\left(\dfrac{9}{2}\right)^2-3\left(\dfrac{9}{2}\right)-10\right)=0 \cdot \left(\left(\dfrac{9}{2}\right)^2-3\left(\dfrac{9}{2}\right)-10\right)=0
- \left(4\left(-\dfrac{9}{2}\right)^2-81\right)\left(\left(-\dfrac{9}{2}\right)^2-3\left(-\dfrac{9}{2}\right)-10\right)=0 \cdot \left(\left(-\dfrac{9}{2}\right)^2-3\left(-\dfrac{9}{2}\right)-10\right)=0
- \left(4\left(5\right)^2-81\right)\left(\left(5\right)^2-3\left(5\right)-10\right)=\left(4\left(5\right)^2-81\right)\cdot 0=0
- \left(4\left(-2\right)^2-81\right)\left(\left(-2\right)^2-3\left(-2\right)-10\right)=\left(4\left(-2\right)^2-81\right) \cdot 0=0
This confirms that, x=\dfrac{9}{2}, x=-\dfrac{9}{2},x=5,x=-2 are the solutions of \left(4x^2-81\right)\left(x^2-3x-10\right)=0.
Example 2
Solve the equation 4x^5-9x^3=72-32x^2.
Approach
We need to first determine whether the equation is in standard form. If it is not, we need to convert it into standard form before we apply some factoring techniques to obtain the solutions.
Solution
We can see that the equation 4x^5-9x^3=72-32x^2 is not in standard form because it is not equal to 0. We can convert the equation to standard form by moving the terms 72 and -32x^2 to the left-hand side of the equation using inverse operations. This leaves 0 on the right-hand side of the equation. The equation in standard form is 4x^5-9x^3+32x^2-72=0
We can proceed by factoring the left-hand side by grouping
| \displaystyle 4x^5-9x^3+32x^2-72 | \displaystyle = | \displaystyle 0 | Standard form |
| \displaystyle (4x^5+32x^2)+(-9x^3-72) | \displaystyle = | \displaystyle 0 | Group terms based on common factors |
| \displaystyle 4x^2(x^3+8)-9(x^3+8) | \displaystyle = | \displaystyle 0 | Factor out the GCF: 4x^2 and -9 |
| \displaystyle \left(4x^2-9\right)\left(x^3+8\right) | \displaystyle = | \displaystyle 0 | Factor out the common binomial |
Observe that 4x^2-9=(2x)^2-(3)^2. This means that 4x^2-9 is a difference of two squares.
| \displaystyle a^2-b^2 | \displaystyle = | \displaystyle (a-b)(a+b) | Formula for difference of two squares |
| \displaystyle (2x)^2-(3)^2 | \displaystyle = | \displaystyle (2x-3)(2x+3) | Substitute a=2x and b=3 |
So we have 4x^2-9=(2x-3)(2x+3).
Also, observe that x^3+8=(x)^3+(2)^3. This means that x^3+8 is a sum of two cubes.
| \displaystyle a^3+b^3 | \displaystyle = | \displaystyle (a+b)\left(a^2-ab+b^2\right) | Formula for sum of two cubes |
| \displaystyle (x)^3+(2)^3 | \displaystyle = | \displaystyle (x+2)\left((x)^2-2x+(2)^2\right) | Substitute a=2x and b=3 |
| \displaystyle (x)^3+(2)^3 | \displaystyle = | \displaystyle (x+2)\left(x^2-2x+4\right) | Simplify |
So we have x^3+8=(x+2)\left(x^2-2x+4\right).
The equation 4x^5-9x^3+32x^2-72=0 becomes (2x-3)(2x+3)(x+2)\left(x^2-2x+4\right)=0
The Zero Product Property states that if A \cdot B \cdot C \cdot D=0, then A =0, B=0, C=0 or D=0.
So we get the following equations:2x-3=0, 2x+3=0,x+2=0 \text{ or }x^2-2x+4=0 Solving the first three equations for x, we obtain the following solutions:
2x-3=0 \to x=\dfrac{3}{2}, 2x+3=0 \to x=-\dfrac{3}{2},x+2=0 \to x=-2
Since the expression x^2-2x+4 is not factorable, we will use quadratic formula in solving x^2-2x+4=0:
| \displaystyle x | \displaystyle = | \displaystyle \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} | Quadratic formula |
| \displaystyle x | \displaystyle = | \displaystyle \dfrac{-(-2)\pm\sqrt{\left(-2\right)^2-4(1)(4)}}{2(1)} | Substitute a=1,b=-2 and c=4 |
| \displaystyle x | \displaystyle = | \displaystyle 1\pm \sqrt{3}i | Simplify |
Therefore, the solutions are:x=\dfrac{3}{2},x=-\dfrac{3}{2},x=-2,x=1+ \sqrt{3}i,x=1- \sqrt{3}i
Reflection
We can verify the answer by substituting each solution to 4x^5-9x^3=72-32x^2 or \\4x^5-9x^3+32x^2-72=0.
- 4\left(\dfrac{3}{2}\right)^5-9\left(\dfrac{3}{2}\right)^3+32\left(\dfrac{3}{2}\right)^2-72=0
- 4\left(-\dfrac{3}{2}\right)^5-9\left(-\dfrac{3}{2}\right)^3+32\left(-\dfrac{3}{2}\right)^2-72=0
- 4\left(-2\right)^5-9\left(-2\right)^3+32\left(-2\right)^2-72=0
- 4\left(1+ \sqrt{3}i\right)^5-9\left(1+ \sqrt{3}i\right)^3+32\left(1+ \sqrt{3}i\right)^2-72=0
- 4\left(1-\sqrt{3}i\right)^5-9\left(1-\sqrt{3}i\right)^3+32\left(1-\sqrt{3}i\right)^2-72=0
All the equations are true. Therefore, x=\dfrac{3}{2},x=-\dfrac{3}{2},x=-2,x=1+ \sqrt{3}i,x=1- \sqrt{3}i are the solutions of 4x^5-9x^3=72-32x^2.
Example 3
A toy car manufacturer produces x units per month. The monthly cost for the toy car is given by \\C(x) = 4 x^{2}+5x The monthly gross profit is given by G(x)=x^{3}+3x^2-8x+7
The cost C(x) and gross profit G(x) are both in dollars.
Form an expression for P(x), the net profit after producing and selling x units.
Approach
To find the expression for P(x), we first determine the relationship of the net profit to the gross profit and cost.
Solution
The net profit is obtained by subtracting the cost from the gross profit. That is, P(x)=G(x)-C(x) By substituting the given expressions we get the equation: P(x)=\left(x^{3}+3x^2-8x+7\right)-\left(4 x^{2}+5x\right)=x^3-x^2-13x+7
Therefore, P(x)=x^3-x^2-13x+7
Find the number of units that must be sold in order to make a profit of \$2962.
Approach
We first substitute the profit into the net profit equation, P\left(x\right), obtained in part (a). Then we solve the new equation for the positive real value of x since the number of units can never take negative values.
Solution
The net profit equation is given by P(x)=x^3-x^2-13x+7
To find the number of units that must be sold to earn a profit of \$2962, we set P(x)=2962 and solve for x. So we have x^3-x^2-13x+7=2962
We convert the equation into standard form by subtracting 2962 from both sides of the equation. x^3-x^2-13x-2955=0
Using the Rational Roots Theorem, we can identify potential roots of the equation in the form \dfrac{p}{q} where p is a factor of the constant term -2955 and q is a factor of the leading coefficient 1.
The possible values of p are \pm1,\pm3,\pm5,\pm15,\pm197,\pm591,\pm985, and \pm2955. And the possible values for q are \pm1, therefore the possible roots, \dfrac{p}{q}, are \pm1,\pm3,\pm5,\pm15,\pm197,\pm591,\pm985, and \pm2955.
Recall the Factor Theorem states that, if P\left(c\right)=0 then \left(x-c\right) is a factor of P\left(x\right). So we can text each of the possible roots by substituting them into the equation for x. If the equation evaluates to 0 then we have found a root and therefore a factor of P\left(x\right).
Substituting 15 into the equation for x gives:
| \displaystyle \left(15\right)^3-\left(15\right)^2-13\left(15\right)-2955 | \displaystyle = | \displaystyle 0 | Substitution |
| \displaystyle 0 | \displaystyle = | \displaystyle 0 | Evaluate |
This shows that 15 is a root of the equation and therefore \left(x-15\right). Is a factor of the equation.
Using long or synthetic division, we can express the left-hand side of the equation as a product of two factors. That is, (x-15)\left(x^2+14x+197\right)=0
The Zero Product Property states that A\cdot B=0 implies A=0 or B=0. This means that we can split the equation into two equations. x-15=0 \text{ or }x^2+14x+197=0
x-15=0 gives the solution x=15.
Since x^2+14x+197=0 is not factorable, we use quadratic formula to solve for the solution of x^2+14x+197=0.
| \displaystyle x | \displaystyle = | \displaystyle \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} | Quadratic formula |
| \displaystyle x | \displaystyle = | \displaystyle \dfrac{-14\pm\sqrt{\left(14\right)^2-4(1)(197)}}{2(1)} | Substitute a=1,b=14 and c=197 |
| \displaystyle x | \displaystyle = | \displaystyle -7\pm 2\sqrt{37}i | Simplify |
However, because -7+ 2\sqrt{37}i and -7- 2\sqrt{37}i are complex solutions, we can see that only the solution x=15 is viable since it is a positive real solution and x represents the number of units sold.
Therefore, 15 units of toy cars must be sold to make a profit of \$2962.
Outcomes
MA.912.AR.1.8
Rewrite a polynomial expression as a product of polynomials over the real or complex number system.
MA.912.AR.6.1
Given a mathematical or real-world context, when suitable factorization is possible, solve one-variable polynomial equations of degree 3 or higher over the real and complex number systems.