Grade 12 · MHF4U · Advanced Functions

MHF4U Polynomial Equations Practice Test (with Solutions)

Q: What is on an MHF4U polynomial equations test?

Dividing polynomials, the remainder and factor theorems, factoring cubics and quartics, solving polynomial equations and inequalities, and finding an equation from its zeros and a point. The questions on this page hit each of those.

Q: What is the remainder theorem?

When a polynomial P(x) is divided by x - a, the remainder is P(a). So you can find a remainder without dividing at all: substitute and simplify.

Q: How do you use the factor theorem to factor a cubic?

Test the values that could be zeros: plus or minus a factor of the constant term over a factor of the leading coefficient. When P(a) = 0, x - a is a factor. Divide it out, then factor the quadratic that is left.

Q: How do you solve a polynomial inequality?

Move everything to one side, factor fully, and mark each zero on a number line. Test the sign in each interval, keep the intervals that match the inequality, and check whether the endpoints are included.

Polynomial Equations is where you stop describing polynomials and start solving them. Your test will want long division, the remainder and factor theorems, factoring cubics and quartics, and polynomial inequalities on a number line. Work each question below on paper first, then open the solution.

Free Polynomial Equations Practice Test

8 questions, exactly as a free account gets them in the first Polynomial Equations test on Martian Lab. Try each one on paper first, then open the answer.

Fully Factor. $x^{3} + 8 =$
Show answer
$x^{3}+8=(x+2)(x^{2}-2x+4)$
Divide $8x^{3} - 10x^{2} + x - 5$ by $2x - 3$ then relate $8x^{3} - 10x^{2} + x - 5$ and $2x - 3$ in two different ways (i.e. using a division statement and using quotient form). The division statement is: $\rule{0.5cm}{0.4pt}$ The quotient form is: $\rule{0.5cm}{0.4pt}$
Show solution
See full solution in image below. The division statement is: $8x^{3} - 10x^{2} + x - 5=(4x^{2}+x+2)(2x-3)+1$ The quotient form is: $\dfrac{8x^{3} - 10x^{2} + x - 5}{2x - 3}=4x^{2}+x+2+\dfrac{1}{2x-3}$
Determine the remainder when $2x^{4} + 3x^{3} + 2x^{2} - 5x - 6$ is divided by $x + 4$ . DO NOT DIVIDE.
Show answer
$f(x)=2x^{4}+3x^{3}+2x^{2}-5x-6$ $f(-4)=2(-4)^{4}+3(-4)^{3}+2(-4)^{2}-5(-4)-6$ $f(-4)=366$ $\therefore$ the remainder when $f(x)$ is divided by $x+4$ is $366$
Fully Factor: $5x^{3} - 11x^{2} - 13x + 3$ .
Show answer
$f(x)=5x^{3}-11x^{2}-13x+3$ $f(-1)=5(-1)^{3}-11(-1)^{2}-13(-1)+3=0$ $(x+1)$ is a factor. $(x+1)(5x^{2}-16x+3)=(x+1)(5x-1)(x-3)$
Solve $2x^{3} + 3x^{2} - 50x = 75$ , $x \in R$ , by factoring.
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$2x^{3}+3x^{2}-50x-75=0$ $x^{2}(2x+3)-25(2x+3)=0$ $(x^{2}-25)(2x+3)=0$ $(x+5)(x-5)(2x+3)=0$ $\therefore \; x=-5$ or $x=5$ or $x=-\dfrac{3}{2}$
Solve $3x^{4} + x^{3} - 12x^{2} - 4x < 0$ for $x \in R$ , algebraically. Show your solution using a Real Number Line.
Show solution
$x(3x^{3}+x^{2}-12x-4)<0$ $x(x^{2}(3x+1)-4(3x+1))<0$ $x(x^{2}-4)(3x+1)<0$ $x(x-2)(x+2)(3x+1)<0$ $x=0,\;2,\;-2,\;-\dfrac{1}{3}$ $\therefore \; f(x) < 0$ when $x\in\left(-2,-\dfrac{1}{3}\right)\cup(0,2)$ $Real number line with open endpoints at $-2$, $-\frac{1}{3}$, and $2$, and a marked point at $0$$
Find the equation of the quartic function in STANDARD FORM, with zeros $\pm 1, 2 \pm \sqrt{3}$ and passes through the point $(2, 5)$ .
Show solution
$f(x)=a(x-1)(x+1)\left[x-(2+\sqrt{3})\right]\left[x-(2-\sqrt{3})\right]$ $=a(x^{2}-1)(x-2-\sqrt{3})(x-2+\sqrt{3})$ $=a(x^{2}-1)\left[(x-2)-\sqrt{3}\right]\left[(x-2)+\sqrt{3}\right]$ $=a(x^{2}-1)\left[(x-2)^{2}-3\right]$ $=a(x^{2}-1)(x^{2}-4x+1)$ $=a(x^{4}-4x^{3}+4x-1)$ $5=a(16-32+8-1)$ $a=-\dfrac{5}{9}$ $f(x)=-\dfrac{5}{9}x^{4}+\dfrac{20}{9}x^{3}-\dfrac{20}{9}x+\dfrac{5}{9}$
When the polynomial $ax^3 + 3x^2 + cx + 3$ is divided by $x + 2$ , the remainder is 12. If $x - 1$ is a factor of the polynomial, determine the values of $a$ and $c$ .
Show solution
$12=a(-2)^{3}+3(-2)^{2}+c(-2)+3$ $12=-8a+12-2c+3$ $-3=-8a-2c$ (1) $0=a(1)^{3}+3(1)^{2}+c(1)+3$ $0=a+3+c+3$ $-6=a+c$ $c=-6-a$ (2) sub (2) into (1) $-3=-8a-2(-6-a)$ $-3=-8a+12+2a$ $-15=-6a$ $a=\frac{5}{2}$ sub $a$ into (2) $c=-6-\frac{5}{2}$ $c=-\frac{17}{2}$ $\therefore \; a=\frac{5}{2}$ and $c=-\frac{17}{2}$

This Unit on Martian Lab

The lessons go from dividing polynomials through the factor and remainder theorems to inequalities and word problems.

Polynomial Operations
Factor and Remainder Theorem
Factor Theorem and Rational Zeros
Polynomial Inequalities
Polynomial Word Problems

Then 4 full Polynomial Equations practice tests to check yourself.

Start Polynomial Equations free

Polynomial Equations FAQ

What is on an MHF4U polynomial equations test?+

Dividing polynomials, the remainder and factor theorems, factoring cubics and quartics, solving polynomial equations and inequalities, and finding an equation from its zeros and a point. The questions on this page hit each of those.

What is the remainder theorem?+

When a polynomial P(x) is divided by x - a, the remainder is P(a). So you can find a remainder without dividing at all: substitute and simplify.

How do you use the factor theorem to factor a cubic?+

Test the values that could be zeros: plus or minus a factor of the constant term over a factor of the leading coefficient. When P(a) = 0, x - a is a factor. Divide it out, then factor the quadratic that is left.

How do you solve a polynomial inequality?+

Move everything to one side, factor fully, and mark each zero on a number line. Test the sign in each interval, keep the intervals that match the inequality, and check whether the endpoints are included.

Polynomial Equations Topics

One idea at a time: a short explanation, then five worked examples.

Long and Synthetic Division

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