Polynomials - High School Math

Card 0 of 7

Question

Find the zeros of the following polynomial:

$f(x) = x^{4} - 4x^{3} - 7x^{2} + 22x + 24$

Answer

First, we need to find all the possible rational roots of the polynomial using the Rational Roots Theorem:

$\frac{\textup{factors of the constant}}{\textup{factors of the leading coefficient}}$

Since the leading coefficient is just 1, we have the following possible (rational) roots to try:

±1, ±2, ±3, ±4, ±6, ±12, ±24

When we substitute one of these numbers for , we're hoping that the equation ends up equaling zero. Let's see if is a zero:

$f(-1)=(-1)^{4}-4(-1)^{3}-7(-1)^{2}+22(-1)+24$

Since the function equals zero when is , one of the factors of the polynomial is . This doesn't help us find the other factors, however. We can use synthetic substitution as a shorter way than long division to factor the equation.

$\textup{-1 } | \textup{ 1 -4 -7 22 24}$

$\small \textup{ -1 }\textup{ 5 }\textup{ 2 }\textup{ -24}$

$\frac{ }{\textup{ 1 -5 -2 24 }\textup{ 0}}$

Now we can factor the function this way:

$f(x)=(x+1)(x^{3}-5x^{2}-2x+24)$

We repeat this process, using the Rational Roots Theorem with the second term to find a possible zero. Let's try :

$f(-2)=[(-2)+1][(-2)^{3}-5(-2)^{2}-2(-2)+24]$

When we factor using synthetic substitution for , we get the following result:

$f(x)=(x+1)(x+2)(x^{2}-7x+12)$

Using our quadratic factoring rules, we can factor completely:

Thus, the zeroes of are

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Question

Simplify the following polynomial function:

$ab^{-2}(a^{2}b + a^{-1}b^{3}-a^{3}b^{-1})$

Answer

First, multiply the outside term with each term within the parentheses:

$ab^{-2}(a^{2}b + a^{-1}b^{3}-a^{3}b^{-1})$

$a^{3}b^{-1} + b-a^{4}b^{-3}$

Rearranging the polynomial into fractional form, we get:

$\frac{a^{3}}{b}+b-\frac{a^{4}}{b^{3}}$

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Question

Simplify the following polynomial:

$(\frac{-a^{-2}b^{3}c^{-1}}{3a^{-4}b^{-1}c^{3}})^{-2}$

Answer

To simplify the polynomial, begin by combining like terms:

$(\frac{-a^{-2}b^{3}c^{-1}}{3a^{-4}b^{-1}c^{3}})^{-2}$

$(\frac{-a^{2}b^{4}}{3c^{4}})^{-2}$

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Question

Factor the polynomial if the expression is equal to zero when $x=-6, -4,\ or\ 2$ .

Answer

Knowing the zeroes makes it relatively easy to factor the polynomial.

The expression fits the description of the zeroes.

Now we need to check the answer.

$\small (x^2+2x-8)(x+6)=x^3+6x^2+2x^2+12x-8x-48=x^3+8x^2+4x-48$

We are able to get back to the original expression, meaning that the answer is .

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Question

A polyomial with leading term $x^ {3}$ has 5 and 7 as roots; 7 is a double root. What is this polynomial?

Answer

Since 5 is a single root and 7 is a double root, and the degree of the polynomial is 3, the polynomial is $(x-5)(x-7)^{2}$ . To put this in expanded form:

$(x-7)^{2} = x^{2} - 2 \cdotx\cdot 7 + 7^{2} = x^{2} - 14x + 49$

$(x-5)(x-7)^{2} =(x-5) \left ( x^{2} - 14x + 49 \right )$

$=x \left ( x^{2} - 14x + 49 \right ) -5\left( x^{2} - 14x + 4 9 \right )$

$= x^{3} - 14x^{2} + 49x - \left( 5x^{2} - 70x + 245 \right )$

$= x^{3} - 14x^{2} + 49x - 5x^{2} + 70x -245$

$= x^{3} - 19x^{2} + 119x -245$

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Question

A polyomial with leading term $x^{3}$ has 6 as a triple root. What is this polynomial?

Answer

Since 6 is a triple root, and the degree of the polynomial is 3, the polynomial is $(x-6)^{3}$ , which we can expland using the cube of a binomial pattern.

$(x-6)^{3} =x ^{3} - 3\cdot x^{2} \cdot 6 + 3\cdot x\cdot 6 ^{2} -6^{3}$

$x ^{3} - 18 x^{2}+ 108x -216$

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Question

What are the solutions to $y = x^{2} + x -6$ ?

Answer

When we are looking for the solutions of a quadratic, or the zeroes, we are looking for the values of such that the output will be zero. Thus, we first factor the equation.

$0 = x^{2} + x - 6 = (x+3)(x-2)$

Then, we are looking for the values where each of these factors are equal to zero.

implies

and implies

Thus, these are our solutions.

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