To find \(\displaystyle b(3x-1)\), substitute \(\displaystyle (3x-1)\) for \(\displaystyle x\) in the original equation:

\(\displaystyle 2(3x-1)^2+(3x-1)-1\).

Use FOIL or the Square of Binomial Rule to find \(\displaystyle (3x-1)^2=(9x^2-6x+1)\).

Recall that FOIL stands for the multiplication between the First components in both binomials followed by the Outer components, then the Inner components, and lastly the Last components.

Therefore, \(\displaystyle b(3x-1)=2(9x^2-6x+1)+(3x-1)-1\). 

You can then simplify the equation.

Distribute the multiplier: \(\displaystyle 18x^2-12x+2+3x-1-1\)

Combine like terms: \(\displaystyle b(3x-1)=18x^2-9x\).

To find \(\displaystyle 3b(x)\), distribute 3 throughout the equation to get:

\(\displaystyle 3b(x)=3(2x^2+x-1)=\)\(\displaystyle 6x^2+3x-3\). 

Subtract the two expressions:

\(\displaystyle (18x^2-9x)-(6x^2+3x-3)\)

\(\displaystyle =12x^2-12x+3.\)

For \(\displaystyle f(5a-2)\), substitue\(\displaystyle (5a-2)\) for \(\displaystyle x\):

\(\displaystyle (5a-2)^2-5(5a-2)+8\).

Use FOIL or square of a binomial to find \(\displaystyle (5a-2)^2=25a^2-20a+4\).

Recall that FOIL stands for the multiplication between the First components in both binomials followed by the Outer components, then the Inner components, and lastly the Last components.

Therefore, \(\displaystyle f(5a-2)=(25a^2-20a+4)-5(5a-2)+8\)

Distribute and combine like terms to simplify:

\(\displaystyle (25a^2-20a+4)-(25a-10)+8\)

\(\displaystyle f(5a-2)=25a^2-45a+22\).

For \(\displaystyle 3f(2a)\), first substitute \(\displaystyle (2a)\) for \(\displaystyle x\):

\(\displaystyle 2a^2-5(2a)+8\).

Multiply the entire expression by 3:

\(\displaystyle 3f(2a)=12a^2-30a+24\).

Add both expressions:

\(\displaystyle (25a^2-45a+22)+(12a^2-30a+24)\)

\(\displaystyle =37a^2-75a+46\)

To solve this problem, plug in 2p for x in the function: \(\displaystyle 2(2p)^3\). Then, simplify: \(\displaystyle 2(8p^3)=16p^3\).

The question asks us to put the expression of \(\displaystyle g(x)\) into the expression for \(\displaystyle f(x)\) anyplace there is an \(\displaystyle x\):

\(\displaystyle f(g(x))=(-3x^{-\frac{1}{2}})^2+1\)

The 2nd power needs to be distributed to both the \(\displaystyle -3\) and \(\displaystyle x^{-\frac{1}{2}}\). The first term then becomes:

\(\displaystyle 9x^{-1} = \frac{9}{x}\)

The final answer is then \(\displaystyle \frac{9}{x}+1\)

To solve this problem, simply plug in 1 wherever you see x.

\(\displaystyle -2(1^3)+1-4=-2+1-4=-5\).

Therefore,

\(\displaystyle f(1)=-5\).

To determine the end behavior for a function, we must look at the degree and the sign associated with the function. For this function: 

\(\displaystyle f(x)=x^6-7x^3+2x^2-5\)

The degree is 6 (the highest power) which is an even number, and the sign is positive (the sign associated with the leading coefficient, which in this case is positive 1). 

Taken together, we can see that we have an "even positive" function. Even functions always go the same direction at both extreme ends. Think of a parabola, both sides either both go up or both go down - this is the same for all even functions. 

Since this is an even positive; both sides will approach positive infinity. 

We express this mathematically when we say that as x approaches negative infinity (left side) the function will approach positive infinity:

\(\displaystyle As\ x\rightarrow\ -\infty; f(x)\rightarrow\infty;\)

....and as x approaches positive infinity (right side) the function again approaches positive infinity. 

\(\displaystyle As\ x\rightarrow\infty; f(x)\rightarrow\infty;\)

Rewrite this equation in order of high to lower powers.

\(\displaystyle y=x^3-4x^2-21x\)

Factor out an x-term  from the equation.  The equation becomes:

\(\displaystyle y=x(x^2-4x-21)\)

Factorize the term inside the parentheses.

\(\displaystyle y=x(x-7)(x+3)\)

Set each individual term equal to zero and solve for \(\displaystyle x\).

\(\displaystyle x=0\)

\(\displaystyle x-7=0, x=7\)

\(\displaystyle x+3=0, x=-3\)

The zeros are:  \(\displaystyle -3,0,7\)

One of the possible root is:  \(\displaystyle -3\)

The vertex is the minimum or maximum of a parabola.

Write the vertex formula for the polynomial \(\displaystyle y=ax^2+bx+c\).

\(\displaystyle x=-\frac{b}{2a}\)

\(\displaystyle y=-3x^2+20x-3\)

\(\displaystyle a=-3, b=20\)

Substitute the coefficients.

\(\displaystyle x=-\frac{b}{2a}=-\frac{20}{2(-3)} = \frac{10}{3}\)

The answer is:  \(\displaystyle \frac{10}{3}\)

In order to factorize this quadratic, we will need to identify the roots of the first and last term and order it into the two binomials.

We know that it will be in the form of:

\(\displaystyle (Ax+ B)(Cx+D)\)

The value of \(\displaystyle 7x^2\) can be divided into \(\displaystyle (7x)(x)\), and is the only possibility to be replaced with \(\displaystyle A\) and \(\displaystyle C\).

\(\displaystyle Ax=7x , Cx=x\)

Substitute this into the binomials.

\(\displaystyle (7x+ B)(x+D)\)

Now we need to determine \(\displaystyle B\times D\) such that it will equal to 12, and satisfy the central term of \(\displaystyle -44x\).

The roots of 12 that can be interchangeable are:

\(\displaystyle [1,12], [2,6],[3,4]\) 

The only terms that are possible are \(\displaystyle [2,6]\) since

\(\displaystyle (Ax)(D)+(B)(Cx) = 7x(-6)+(-2)(x) = -44x\).

Remember that we must have a positive ending term!

This means that \(\displaystyle B=-2, D=-6\).  

Substitute the terms.

The answer is:  \(\displaystyle (7x-2)(x-6)\)

By the Rational Zeroes Theorem (RZT), if a polynomial has only integer coefficients, then any rational zero must be the positive or negative quotient of a factor of the constant and a factor of the coefficient of greatest degree. In other words, the RZT only gives a set of rational numbers of which any rational zeroes must be an element; it does not dictate what rational zeroes, if any, must actually be zeroes of the polynomial.

For example, examine the polynomial 

\(\displaystyle P(x) = 3x^{7} + 20\),

which satisfies the given characteristics in that its first and last terms are \(\displaystyle 3x^{7}\) and \(\displaystyle 20\). If \(\displaystyle \frac{20}{3}\) is a zero, then 

\(\displaystyle P \left ( \frac{20}{3} \right ) = 0\).

Substitute \(\displaystyle \frac{20}{3}\) for \(\displaystyle x\):

\(\displaystyle P \left ( \frac{20}{3} \right ) = 3 \left ( \frac{20}{7} \right )^{7} + 20 \approx 1,574.3\)

\(\displaystyle \frac{20}{3}\) is not a zero of \(\displaystyle P(x) = 3x^{7} + 20\).

Therefore, it does not hold in general that a polynomial with the given highest- and lowest-degree terms has \(\displaystyle \frac{20}{3}\) as a zero.