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53358711
The graph shows the functions \(f\), \(g\), and their sum \(h\), where \(h(x)=f(x)+g(x)\). Decide whether each statement is true or false. Justify your decisions. a) If \(f\) has a zero at some x-value, then \(h\) also has a zero there. b) At every x-value where the graphs of \(f\) and \(g\) intersect, \(h(x)\) is twice \(f(x)\). c) If \(f(x)=-g(x)\), then \(h\) has a zero at that x-value.
Figure for problem 533587

Hints

- Use a specific zero of \(f\) from the graph to test part a. - At an intersection, the two functions have equal outputs. - Substitute \(f(x)=-g(x)\) into the definition of \(h\).

Solution

1. Statement a is false. For example, \(f(-2)=0\), but \(g(-2)=4\). Therefore, \(h(-2)=0+4=4\ne0\). 2. Statement b is true. At an intersection, \(f(x)=g(x)\). Therefore, \(h(x)=f(x)+g(x)=2f(x)\). 3. Statement c is true. If \(f(x)=-g(x)\), then \(h(x)=f(x)+g(x)=-g(x)+g(x)=0\).

Answer

a) False b) True c) True
53359411
Let \(f(x)=x^2\) and \(g(x)=2x\). Each panel shows the graphs of \(f\) and \(g\), along with one candidate graph labeled p or q. One candidate represents the sum \(s(x)=f(x)+g(x)\), and the other represents the difference \(d(x)=f(x)-g(x)\). Match p and q to the correct formulas. Justify your answer by testing a point or analyzing the zeros.
Figure for problem 533594

Hints

- Choose a simple input such as \(x=1\) and evaluate both combined functions. - Compare those outputs with the candidate graphs. - Alternatively, factor each combined polynomial and compare its zeros with the graphs.

Solution

1. Test \(x=1\): \(s(1)=1^2+2\cdot 1=3\), while \(d(1)=1^2-2\cdot 1=-1\). 2. Graph p passes through \((1, 3)\), so p represents \(s(x)=x^2+2x\). 3. Graph q passes through \((1, -1)\), so q represents \(d(x)=x^2-2x\). 4. The zeros confirm the matches: \(s(x)=x(x+2)\) has zeros \(-2\) and \(0\), while \(d(x)=x(x-2)\) has zeros \(0\) and \(2\).

Answer

p: \(s(x)=x^2+2x\) q: \(d(x)=x^2-2x\)
53359511
The panels show the graphs of \(f(x)=x^2\) and \(g(x)=x^4\). The graphs labeled p and q were formed by subtracting these functions. Which graph represents \(h(x)=f(x)-g(x)\), and which represents \(k(x)=g(x)-f(x)\)? Justify your answer by comparing \(x^2\) and \(x^4\) on \((-1, 1)\).
Figure for problem 533595

Hints

- Compare \(x^2\) and \(x^4\) when \(|x|<1\). - Determine the sign of each difference on that interval. - Notice that the two difference functions are opposites.

Solution

1. For \(-1<x<1\), \(x^4<x^2\) except at \(x=0\), where they are equal. 2. Therefore, \(h(x)=x^2-x^4\) is positive on \((-1, 0)\cup(0, 1)\) and equals zero at \(x=0\). Graph q has this behavior. 3. Also, \(k(x)=x^4-x^2=-h(x)\), so it is negative on \((-1, 0)\cup(0, 1)\). Graph p has this behavior.

Answer

p: \(k(x)=x^4-x^2\) q: \(h(x)=x^2-x^4\)
53359611
Let \(u\) and \(v\) be functions, and define their product by \(w(x)=u(x)v(x)\), with \(\operatorname{dom}(w)=\operatorname{dom}(u)\cap\operatorname{dom}(v)\). The graph shows one example. Suppose \(a\in\operatorname{dom}(w)\). Decide whether this statement is always true, and justify your answer from the definition of \(w\): “If \(u(a)=0\), then \(w(a)=0\).”
Figure for problem 533596

Hints

- Translate “\(u\) has a zero at \(a\)” into an equation. - Substitute that equation into the definition \(w(a)=u(a)v(a)\). - Explain why the domain condition matters.

Solution

1. Since \(a\in\operatorname{dom}(w)\), both \(u(a)\) and \(v(a)\) are defined. 2. By the definition of the product function, \(w(a)=u(a)v(a)\). 3. If \(u(a)=0\), then \(w(a)=0\cdot v(a)=0\). 4. Therefore, the statement is always true under the stated domain condition.

Answer

True. If \(a\in\operatorname{dom}(u)\cap\operatorname{dom}(v)\) and \(u(a)=0\), then \(w(a)=u(a)v(a)=0\).
53359711
Let \(h(x)=f(x)-g(x)\). The graph shows one example of \(f\), \(g\), and \(h\). Decide whether this statement is always true: “Whenever the graph of \(f\) is above the graph of \(g\), the corresponding value of \(h\) is positive.” Justify your answer algebraically.
Figure for problem 533597

Hints

- Translate the relative positions of the graphs into an inequality. - Subtract the same quantity from both sides. - Use the definition of \(h\).

Solution

1. If the graph of \(f\) is above the graph of \(g\) at an input \(x\), then \(f(x)>g(x)\). 2. Subtract \(g(x)\) from both sides to obtain \(f(x)-g(x)>0\). 3. Since \(h(x)=f(x)-g(x)\), it follows that \(h(x)>0\). Therefore, the statement is always true.

Answer

True. The condition \(f(x)>g(x)\) is equivalent to \(f(x)-g(x)>0\), so \(h(x)>0\).
52878111
The polynomial function \(f\) is given in factored form by \(f(x) = 3(x - 1)(x + 2)(x - 4)\). 1. Multiply the factors step by step and write the function in standard form \(f(x) = a_3x^3 + a_2x^2 + a_1x + a_0\). 2. Compare the value of \(a_3\) with the coefficient outside the factors. Explain why, in general, the leading coefficient of \(a_n(x - x_1)(x - x_2)\cdots(x - x_n)\) is \(a_n\).

Hints

- Use the distributive property to multiply two binomials first. - Multiply the resulting quadratic by the remaining binomial before applying the outside coefficient. - Identify how the highest-degree term is formed from the factors. - Track what the outside coefficient does to the leading term.

Solution

1. Multiply the first two factors: \((x - 1)(x + 2) = x^2 + x - 2\). 2. Multiply by the third factor: \((x^2 + x - 2)(x - 4) = x^3 - 3x^2 - 6x + 8\). 3. Multiply by the outside coefficient: \(f(x) = 3x^3 - 9x^2 - 18x + 24\). 4. Thus, \(a_3 = 3\), which equals the coefficient outside the product. 5. In a product of \(n\) monic linear factors, the term \(x^n\) is produced only by multiplying the \(x\)-term from every factor. Its coefficient is \(1\). Multiplying the entire product by \(a_n\) makes the leading term \(a_nx^n\), so the leading coefficient is \(a_n\).

Answer

1. \(f(x) = 3x^3 - 9x^2 - 18x + 24\) 2. \(a_3 = 3\). In general, multiplying the \(x\)-term from each monic linear factor produces \(x^n\), and the outside coefficient \(a_n\) makes the leading term \(a_nx^n\).
52878211
Consider the function \(p(x) = -2(x + 1)^2(x - 3)\). 1. Expand the expression and write the polynomial in standard form. 2. State the leading coefficient \(a_3\) and the constant term \(a_0\). 3. Explain how to identify the leading coefficient directly from this factored form without expanding the entire expression.

Hints

- Use the square of a binomial to expand \((x + 1)^2\). - The leading coefficient is the coefficient of the highest-degree term. - The constant term contains no variable. - To find only the leading term, multiply only the highest-degree term from each factor.

Solution

1. Expand the squared factor: \((x + 1)^2 = x^2 + 2x + 1\). 2. Multiply by the remaining factor: \((x^2 + 2x + 1)(x - 3) = x^3 - x^2 - 5x - 3\). 3. Multiply by \(-2\): \(p(x) = -2x^3 + 2x^2 + 10x + 6\). 4. Therefore, the leading coefficient is \(a_3 = -2\), and the constant term is \(a_0 = 6\). 5. Each linear factor is monic, so multiplying the leading terms gives \(x^2 \cdot x = x^3\). The outside coefficient \(-2\) therefore becomes the coefficient of \(x^3\).

Answer

1. \(p(x) = -2x^3 + 2x^2 + 10x + 6\) 2. \(a_3 = -2\) and \(a_0 = 6\) 3. Because the linear factors are monic, their leading terms multiply to \(x^3\); the outside coefficient \(-2\) is therefore the leading coefficient.
52878611
Consider the function \(f(x) = (x + 2)(x - 1)^2\). a) Expand the expression and write \(f(x)\) in standard polynomial form. b) State the degree and leading coefficient of the function. c) Find all zeros and state the multiplicity of each zero.

Hints

- Expand the squared binomial first. - Standard form lists polynomial terms in descending powers. - The degree and leading coefficient come from the highest-degree term. - The factored form makes the zeros easiest to identify. - The exponent on a factor gives the multiplicity of its zero.

Solution

1. Expand the squared factor: \((x - 1)^2 = x^2 - 2x + 1\). 2. Multiply by \(x + 2\): \((x + 2)(x^2 - 2x + 1) = x^3 - 2x^2 + x + 2x^2 - 4x + 2\). 3. Combine like terms: \(f(x) = x^3 - 3x + 2\). 4. The highest exponent is \(3\), so the degree is \(3\). The coefficient of \(x^3\) is \(1\), so the leading coefficient is \(1\). 5. From the factored form, \(x + 2 = 0\) gives \(x = -2\), and \(x - 1 = 0\) gives \(x = 1\). 6. The factor \(x + 2\) occurs once, so \(x = -2\) has multiplicity 1. The factor \(x - 1\) is squared, so \(x = 1\) has multiplicity 2.

Answer

a) \(f(x) = x^3 - 3x + 2\) b) Degree \(3\); leading coefficient \(1\) c) \(x = -2\) with multiplicity 1; \(x = 1\) with multiplicity 2
53232411
A company has two divisions. Division A's monthly profit or loss, in thousands of dollars, is modeled by \(f(x)=-0.1x^2+0.8x+1\). Division B's monthly profit or loss is modeled by \(g(x)=0.5x-2\), where \(x\) is the month number. The graph shows both functions. Define the company's total monthly profit by \(h(x)=f(x)+g(x)\). a) Write \(h(x)\) as a simplified polynomial. Then find the company's total profit in months \(x=2\) and \(x=8\). b) In which month is \(h(x)\) equal to Division A's profit \(f(x)\)? Explain your answer using \(g(x)\). c) Determine whether this statement is true or false, and justify your answer: “Whenever \(h(x)>0\), both divisions must have earned a profit.”
Figure for problem 532324

Hints

- Add the two polynomial expressions by combining like terms. - If \(f(x)+g(x)=f(x)\), determine what \(g(x)\) must equal. - To disprove the statement, find one month when the sum is positive but one function value is negative.

Solution

1. Add like terms: \(h(x)=(-0.1x^2+0.8x+1)+(0.5x-2)=-0.1x^2+1.3x-1\). Then \(h(2)=-0.4+2.6-1=1.2\), so the total profit is \(\$1200\). Also, \(h(8)=-6.4+10.4-1=3\), so the total profit is \(\$3000\). 2. If \(h(x)=f(x)\), then \(f(x)+g(x)=f(x)\), which requires \(g(x)=0\). Solve \(0.5x-2=0\) to get \(x=4\). 3. The statement is false. At \(x=2\), \(h(2)=1.2>0\), but \(g(2)=-1<0\). The company has a positive total profit even though Division B has a loss.

Answer

a) \(h(x)=-0.1x^2+1.3x-1\); month \(2\): \(\$1200\); month \(8\): \(\$3000\) b) Month \(4\), because \(g(4)=0\) c) False; for example, \(h(2)>0\) while \(g(2)<0\).

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