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2.1: 2.1 Factoring Review

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To factor means to write an expression as a product instead of a sum. Factoring is particularly useful when solving equations set equal to zero because then logically at least one factor must be equal to zero. In PreCalculus, you should be able to factor even when there is no obvious greatest common factor or the difference is not between two perfect squares.

How do you use the difference of perfect squares factoring technique on polynomials that don’t contain perfect squares and why would this be useful?

Factoring Functions

A polynomial is a sum of a finite number of terms. Each term consists of a constant that multiplies a variable. The variable may only be raised to a non-negative exponent. The letters \(a, b, c \ldots\) in the following general polynomial expression stand for regular numbers like \(0,5,-\frac{1}{4}, \sqrt{2}\) and the \(x\) represents the variable.

\(a x^{n}+b x^{n-1}+\ldots+f x^{2}+g x+h\)

You have already learned many properties of polynomials. For example, you know the commutative property which states that terms of a polynomial can be rearranged to create an equivalent polynomial. When two polynomials are added, subtracted or multiplied the result is always a polynomial. This means polynomials are closed under addition, and is one of the properties that makes the factoring of polynomials possible. Polynomials are not closed under division because dividing two polynomials could result in a variable in the denominator, which is a rational expression (not a polynomial).

There are three methods for factoring that are essential to master.

Greatest Common Factor Method

The first method you should always try is to factor out the greatest common factor (GCF) of the expression.

To factor the following expression, first apply the GCF method:

\(-\frac{1}{2} x^{4}+\frac{7}{2} x^{2}-6\)

To find the GCF, it is common to try to factor out the \(a\) value. In this case, try factoring out \(-\frac{1}{2}\)

\(-\frac{1}{2} x^{4}+\frac{7}{2} x^{2}-6=-\frac{1}{2}\left(x^{4}-7 x^{2}+12\right)\)

In order to check to see that this is an equivalent expression, you need to distribute the \(-\frac{1}{2}\). When you distribute, the first coefficient matches because it just gets multiplied by 1 , the second term becomes \(\frac{7}{2}\) and the third term becomes - \(6 .\) Note that this expression is not completely factored yet but it is simplified as much as it can be with just the GCF method.

Factoring Into Binomials Method

The second method you should try is to see if you can factor the expression into the product of two binomials.

To continue factoring the expression from the Greatest Common Factor section, factor the following expression into the product of two binomials and a constant:

\(-\frac{1}{2}\left(x^{4}-7 x^{2}+12\right)\)

Many students familiar with basic factoring may be initially stuck on a problem like this. However, you should recognize that beneath the \(4^{t h}\) degree and the \(-\frac{1}{2}\) the problem boils down to being able to factor \(u^{2}-7 u+12\) which is just \((u-3)(u-4)\).

Start by rewriting the problem: \(-\frac{1}{2}\left(x^{4}-7 x^{2}+12\right)\)

Then choose a temporary substitution: Let \(u=x^{2}\).

Then substitute and factor away. Remember to substitute back at the end.

\(\begin{aligned}-\frac{1}{2}\left(u^{2}-7 u+12\right) &=-\frac{1}{2}(u-3)(u-4) \\ &=-\frac{1}{2}\left(x^{2}-3\right)\left(x^{2}-4\right) \end{aligned}\)

This type of temporary substitution that enables you to see the underlying structure of an expression is very common in calculus. The expression is still not completely factored and since there are no more trinomials, you must apply the last method.

Difference of Squares Method

The third method of basic factoring is the difference of squares. It is recognizable as one square monomial being subtracted from another square monomial.

To finish factoring the resulting expression from the Factoring Into Binomials section, factor the expression into four linear factors and a constant:

\(-\frac{1}{2}\left(x^{2}-3\right)\left(x^{2}-4\right)\)

Many students may recognize that \(x^{2}-4\) immediately factors by the difference of squares method to be \((x-2)(x+2)\). This problem asks for more because sometimes the difference of squares method can be applied to expressions like \(x^{2}-3\) where each term is not a perfect square. The number 3 actually is a square.

\(3=(\sqrt{3})^{2}\)

So the fully factored expression would be:

\(-\frac{1}{2}(x-\sqrt{3})(x+\sqrt{3})(x-2)(x+2)\)

Earlier, you were asked how you use the difference of perfect squares factoring technique on polynomials that don't contain perfect squares and why it would be useful. One reason why it might be useful to completely factor an expression like \(-\frac{1}{2}\left(x^{4}-7 x^{2}+12\right)\) into linear factor is if you wanted to find the roots of the function \(f(x)=-\frac{1}{2}\left(x^{4}-7 x^{2}+12\right)\). The roots are \(x=\pm \sqrt{3},\pm 2\)

You should recognize that \(x^{2}-3\) can still be thought of as the difference of perfect squares because the number 3 can be expressed as \((\sqrt{3})^{2}\). Rewriting the number 3 to fit a factoring pattern that you already know is an example of using the basic factoring techniques at a PreCalculus level.

Factor the following expression into strictly linear factors if possible. If not possible, explain why.

\(\begin{aligned} \frac{x^{5}}{3}-\frac{11 x^{3}}{3}+6 x &=\frac{1}{3} x\left(x^{4}-11 x^{2}+18\right) \\ &=\frac{1}{3} x\left(x^{2}-2\right)\left(x^{2}-9\right) \\ &=\frac{1}{3} x(x+\sqrt{2})(x-\sqrt{2})(x+3)(x-3) \end{aligned}\)

\(-\frac{2}{7} x^{4}+\frac{74}{63} x^{2}-\frac{8}{63}\)

For \(-\frac{2}{7} x^{4}+\frac{74}{63} x^{2}-\frac{8}{63},\) let \(u=x^{2}\).

\(=-\frac{2}{7} u^{2}+\frac{74}{63} u-\frac{8}{63}\) \(=-\frac{2}{7}\left(u^{2}-\frac{37}{9} u+\frac{4}{9}\right)\)

Factoring through fractions like this can be extremely tricky. You must recognize that \(-\frac{1}{9}\) and -4 sum to \(-\frac{37}{9}\) and multiply to \(\frac{4}{9}\).

\(=-\frac{2}{7}\left(u-\frac{1}{9}\right)(u-4)\) \(=-\frac{2}{7}\left(x^{2}-\frac{1}{9}\right)\left(x^{2}-4\right)\) \(=-\frac{2}{7}\left(x-\frac{1}{3}\right)\left(x+\frac{1}{3}\right)(x-2)(x+2)\)

\(x^{4}+x^{2}-72\)

\(x^{4}+x^{2}-72=\left(x^{2}-8\right)\left(x^{2}+9\right)\)

Notice that \(\left(x^{2}-8\right)\) can be written as the difference of perfect squares because \(8=(\sqrt{8})^{2}=(2 \sqrt{2})^{2}\). On the other hand, \(x^{2}+9\) cannot be written as the difference between squares because the \(x^{2}\) and the 9 are being added not subtracted. This polynomial cannot be factored into strictly linear factors.

\(x^{4}+x^{2}-72=(x-2 \sqrt{2})(x+2 \sqrt{2})\left(x^{2}+9\right)\)

Factor each polynomial into strictly linear factors if possible. If not possible, explain why not.

1. \(x^{2}+5 x+6\)

2. \(x^{4}+5 x^{2}+6\)

3. \(x^{4}-16\)

4. \(2 x^{2}-20\)

5. \(3 x^{2}+9 x+6\)

6. \(\frac{x^{4}}{2}-5 x^{2}+\frac{9}{2}\)

7. \(\frac{2 x^{4}}{3}-\frac{34 x^{2}}{3}+\frac{32}{3}\)

8. \(x^{2}-\frac{1}{4}\)

9. \(x^{4}-\frac{37 x^{2}}{4}+\frac{9}{4}\)

10. \(\frac{3}{4} x^{4}-\frac{87}{4} x^{2}+75\)

11. \(\frac{1}{2} x^{4}-\frac{29}{2} x^{2}+50\)

12. \(\frac{x^{4}}{2}-\frac{5 x^{2}}{9}+\frac{1}{18}\)

13. \(x^{4}-\frac{13}{36} x^{2}+\frac{1}{36}\)

14. How does the degree of a polynomial relate to the number of linear factors?

15. If a polynomial does not have strictly linear factors, what does this imply about the type of roots that the polynomial has?

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1.16: Factoring Trinomials and Mixed Factoring

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• Samar ElHitti, Marianna Bonanome, Holly Carley, Thomas Tradler, & Lin Zhou
• CUNY New York City College of Technology & NYC College of Technology via New York City College of Technology at CUNY Academic Works

Factoring Trinomials \(a x^{2}+b x+c\) by the ac-Method

We know that multiplying two binomials by the FOIL method results in a four-term polynomial and in many cases it can be combined into a three-term polynomial. For example:

\[ \begin{align*} (x+3)(2 x+1) &=2 x^{2}+1 x+6 x+3 \\[4pt] &=2 x^{2}+7 x+3 . \end{align*}\]

This indicates that if we want to factor the expression \(2 x^{2}+7 x+3\), we will get a product of two binomials \((x+3)\) and \((2 x+1)\), that is,

\[2 x^{2}+7 x+3=(x+3)(2 x+1). \nonumber\]

In this section, we will learn how to reverse the procedure of FOIL to factor trinomials of the form \(a x^{2}+b x+c \). The procedure is called the ac-Method .

• Step 1. Find the product \(a c\), that is the product of the coefficients of the first and last terms.
• Step 2. Find two integers whose product is \(a c\) and whose sum is \(b\). If such an integer pair cannot be found, then the polynomial cannot be factored out.
• Step 3. Use the two integers found in step 2 to rewrite the term \(b x\) as a sum of two terms.
• Step 4. Factor by the grouping method.

For example: Factor \(2 x^{2}+7 x+3\).

Step \(1 \). The product of \(a c\) is \(2 \cdot 3=6\).

Step 2. We look for two numbers whose product is 6 and whose sum is 7 . We can do this by inspection or by writing all pairs of numbers whose product is 6 and calculate the sum for each pair: \(1+6=7,2+3=5\). So 1 and 6 are the numbers we are looking for.

Step 3. We write \(7 x=1 x+6 x\) so

\[2 x^{2}+7 x+3=2 x^{2}+x+6 x+3\nonumber\]

\[\begin{align*}\begin{aligned} 2 x^{2}+x+6 x+3 &=\left(2 x^{2}+x\right)+(6 x+3) ~~ \text{Factor by Grouping} \\ &=x(2 x+1)+3(2 x+1) \\ &=(x+3)(2 x+1) \end{aligned}\end{align*}\nonumber\]

We can check to see if we factored correctly by distributing our answer. We can use the FOIL method learned previously to check if the factored binomials give us the original trinomial \(2 x^{2}+7 x+3\).

Factor the given polynomial by the ac-Method.

a) \(2 x^{2}+15 x-27\)

• Step 1. The product of \(a c=(2)(-27)=-54\)
• Step 2. Now we need to find two integers whose product is - 54 . We can list all the possibilities:

\[(-1)(54), \quad(-2)(27), \quad(-3)(18), \quad(-6)(9),\nonumber\]

\[(1)(-54), \quad(2)(-27), \quad(3)(-18), \quad(6)(-9)\nonumber\]

and calculate the sum of each pair. Only integers -3 and 18 add up to 15 .

• Step 3. We can rewrite the middle term \(15 x=-3 x+18 x \). So \(2 x^{2}+15 x-\) \(27=2 x^{2}-3 x+18 x-27\)
• Step 4. We factor by grouping.

\[\begin{align*} 2 x^{2}+15 x-27 &=2 x^{2}-3 x+18 x-27 \\ &=\left(2 x^{2}-3 x\right)+(18 x-27) \\ &=x(2 x-3)+9(2 x-3) \\ &=(x+9)(2 x-3) \end{align*}\nonumber\]

b) \(12 x^{2}-11 x+2:\)

Step 1. The product of \(a c=(12)(2)=24\)

Step 2. We need to find two integers whose product is 24 and whose sum is -11 . We list all pairs of factors of 24:

\[(1)(24), \quad(2)(12), \quad(3)(8), \quad(4)(6)\nonumber\]

\[(-1)(-24), \quad(-2)(-12), \quad(-3)(-8), \quad(-4)(-6)\nonumber\]

The pair -3 and -8 will have a sum -11.

Step 3. We rewrite the middle term \(-11 x=(-3 x)+(-8 x)\).

Step 4. Then we can finish the factoring.

\[\begin{align*} 12 x^{2}-11 x+2 &=12 x^{2}-3 x+(-8 x)+2 \\ &=\left(12 x^{2}-3 x\right)+(-8 x+2) \\ &=3 x(4 x-1)+(-2)(4 x-1) \\ &=(3 x-2)(4 x-1) \end{align*}\nonumber\]

Note that when we rewrite the middle term, we write it as a sum (even if the second term is negative). This is in order to be able to group without worrying about subtraction. since otherwise the grouping step would look like this:

\[12 x^{2}-3 x-8 x+2=\left(12 x^{2}-3 x\right)-(8 x-2) \nonumber\]

(note the subtraction of 2). Also, note that we factored out (-2) in the second to the last step. This was in order to make sure that \((4 x-1)\) was a common factor.

c) \(3 x^{2}+4 x-2\)

The product of \(a c=(3)(-2)=-6\), and this number factors as:

\[(-1)(6), \quad(-2)(3), \quad(1)(-6), \quad(2)(-3)\nonumber\]

It is clear that none of pairs in the list will give a sum \(4 \). This means that the polynomial \(3 x^{2}+4 x-2\) cannot be factored into two binomials (using integers). We call it a prime polynomial .

Factoring Trinomials \(x^{2}+b x+c\)

In the special case when \(a=1\), the AC-method still works. For example, to factor \(x^{2}-6 x+5\), we first compute \(a c=(1)(5)=5 \). Then we need to find two numbers whose product is 5 and whose sum is \(-6 \). since \((-1)(-5)=5\) and \((-1)+(-5)=-6\), by the grouping method we have:

\[\begin{align*} x^{2}-6 x+5 &=x^{2}-1 x+(-5 x)+5 \\ &=\left(x^{2}-1 x\right)+(-5 x+5) \\ &=x(x-1)+(-5)(x-1) \\ &=(x-5)(x-1) \end{align*}\nonumber\]

Now let's observe the result. The result has the form \((x+[-5])(x+[-1])\), and the two numbers in the two boxes are just the two numbers we get to rewrite the coefficient of the middle term \(-6\), that is -1 and \(-5 \).

This example shows that to factor \(x^{2}+b x+c\), the grouping method can be simplified. We can directly write out the factored form of the polynomial once we know the two numbers that multiply to \(a c\) and add to \(b\). In other words, \(x^{2}+b x+c\) is factored as \((x+\square)(x+\square)\), the product of the two numbers in the boxes being \(a c=(1)(c)=c\) and the sum of the two numbers in the boxes being \(b\),

Factor the given trinomial.

a) \(x^{2}+7 x+10\):

We need to find two numbers whose product is \(a c=c=10\), and whose sum is \(7 \). Number 10 is a product of the following two numbers:

\[(1)(10), \quad(2)(5), \quad(-1)(-10), \quad(-2)(-5)\nonumber\]

The pair 2 and 5 gives a sum \(7\), therefore the trinomial can be factored as:

\[x^{2}+7 x+10=(x+2)(x+5)\nonumber\]

b) \(t^{2}+4 t-12\):

We need to find two numbers whose product is \(a c=c=-12\) whose sum is 4. The number -12 is a product of the following two numbers:

\[(1)(-12), \quad(2)(-6), \quad(3)(-4)\nonumber\]

\[(-1)(12), \quad(-2)(6), \quad(-3)(4).\nonumber\]

The pair -2 and 6 gives a sum \(4\), therefore the trinomial can be factored as:

\[t^{2}+4 t-12=(t+(-2))(t+6)=(t-2)(t+6)\nonumber\]

c) \(x^{2}-3 x-24\):

We need to find two numbers whose product is \(a c=c=-24\) whose sum is -3. The number -24 can be factored as:

\[(1)(-24), \quad(2)(-12), \quad(3)(-8), \quad(4)(-6)\nonumber\]

\[(-1)(24), \quad(-2)(12), \quad(-3)(8), \quad(-4)(6).\nonumber\]

Since none of the pairs in the list adds up to \(-3\), the trinomial cannot be factored as a product of two binomials. This is a prime polynomial.

Mixed Factoring

So far, we have explained the basic techniques of factoring polynomials. Here is the guideline we can follow to select the right method to factor a given polynomial completely.

• Step 1 . Factor out the GCF from all terms if possible.
• Step 2 . Count the number of terms of the polynomial: if the polynomial has two terms, try the formula of difference of two squares; if the polynomial has three terms, try the AC-method; if the polynomial has four terms, try the grouping method.
• Step 3 . Check to see if the factors themselves can be factored. If the answer is yes, then factor them completely using the methods in step 2 .

Factor the given polynomial completely.

a) \(3 x^{2}-12\):

\[\begin{align*} 3 x^{2}-12 &=3\left(x^{2}-4\right) \quad \text { Factor out the GCF 3} \\ &=3(x+2)(x+(-2)) \quad \text { Factor the difference of two squares } x^2-4 \\ &=3(x+2)(x-2) \end{align*}\nonumber\]

b) \(4 x^{3}-20 x^{2}+24 x\):

\[\begin{align*} 4 x^{3}-20 x^{2}+24 x &=4 x\left(x^{2}-5 x+6\right) \quad \text { Factor out the GCF } 4x \\ &=4 x(x+(-2))(x+(-3)) \quad \text { Factor the trinomial } x^{2}-5 x+6 \\ &=4 x(x-2)(x-3) \end{align*}\nonumber\]

c) \(-10 z^{2}-4 z+6\):

\[\begin{align*} -10 z^{2}-4 z+6 &=-2\left(5 z^{2}+2 z-3\right) \quad \text { Factor out the opposite of the GCF } -2 \\ &=-2\left(5 z^{2}-3 z+5 z-3\right) \quad \text { Factor the trinomial } 5 z^{2}+2 z-3 \\ &=-2\left[\left(5 z^{2}-3 z\right)+(5 z-3)\right] \\ &=-2[z(5 z-3)+1(5 z-3)] \\ &=-2(z+1)(5 z-3) \end{align*}\nonumber\]

d) \(x^{3}-7 x^{2}-4 x+28\)

\[\begin{align*} x^{3}-7 x^{2}-4 x+28 &=\left(x^{3}-7 x^{2}\right)+(-4 x+28) \quad \text { Factor by grouping } \\ &=x^{2}(x-7)+(-4)(x-7) \\ &=\left(x^{2}-4\right)(x-7) \\ &=(x+2)(x+(-2))(x-7) \quad \text { Factor } x^{2}-4 \\ &=(x+2)(x-2)(x-7) \end{align*}\nonumber\]

e) \(30 x^{2}+10 x^{4}-280\)

\[\begin{align*} 30 x^{2}+10 x^{4}-280 &=10 x^{4}+30 x^{2}-280 \quad \text { Reorder in decreasing powers of the variable } \\ &=10\left(x^{4}+3 x^{2}-28\right) \quad \text { Factor out the GCF } 10 \\ &=10\left(y^{2}+3 y-28\right) \quad \text { Write } y=x^{2} \text{ to recognize as a quadratic expression} \\ &=10(y+7)(y-4) \\ &=10\left(x^{2}+7\right)\left(x^{2}-4\right) \quad \text { Replace } y \text{ with } x^2 \text{and look to see if anything can be factor out}\\ &= 10\left(x^{2}+7\right)(x+2)(x-2) \end{align*}\nonumber\]

Exit Problem

Factor completely: \(8 x^{2}-10 x+3\)

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Course: Algebra 1   >   Unit 13

• Factoring quadratics as (x+a)(x+b)
• Factoring quadratics: leading coefficient = 1
• Factoring quadratics as (x+a)(x+b) (example 2)
• More examples of factoring quadratics as (x+a)(x+b)

Factoring quadratics intro

• Factoring quadratics with a common factor
• Factoring completely with a common factor
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To Succeed with AI, Adopt a Beginner’s Mindset

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Letting go of ego and expertise allows for openness and curiosity.

Times of substantial tech progress and change, like the current AI revolution, create fear and anxiety. This often causes leaders to fall back on their ego and emphasize their expertise, closing their minds and negatively impacting their people and organizations. Instead, leaders need to take on a beginner’s mindset of openness and curiosity. This is not easy. The more experienced we are, the more locked-in we tend to become in our ways of thinking and doing things. But it is possible, and with employee stress at record highs, it’s necessary. And research shows that the openness that comes with a beginner’s mindset is a crucial factor in achieving better outcomes. There are a few simple questions you can ask yourself to gauge your leadership style and whether you have a beginner’s mindset.

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• Jacqueline Carter is a senior partner and the North American Director of Potential Project. She has extensive experience working with senior leaders to enable them to achieve better performance while enhancing a more caring culture. She is the coauthor, with Rasmus Hougaard, of Compassionate Leadership: How to Do Hard Things in a Human Way and The Mind of the Leader – How to Lead Yourself, Your People, and Your Organization for Extraordinary Results .
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‘Player Kings’ Review: Ian McKellen’s Juicy Assignment as Falstaff

In Robert Icke’s adaptation of Parts 1 and 2 of “Henry IV,” the veteran stage actor’s performance belies his age.

By Matt Wolf

The critic Matt Wolf saw the show in London.

There are two shows for the price of one at “ Player Kings ,” in which the director Robert Icke has combined both of Shakespeare’s “Henry IV” history plays into a self-contained whole.

The production offers a compressed version of the royal accession story that, in this version, runs nearly four hours. It is an opportunity to experience Ian McKellen’s unbridled love of performance. At 84, the production’s leading man possesses an energy and vigor that belie his years.

“ Player Kings ” — which runs at the Noël Coward Theater through June 22, before touring England — is the latest in a wave of recent high-profile Shakespeare productions in London. Uniquely among the other great British theater actors of his generation, McKellen still returns year after year to the stage, recently tackling Lear for a second time and playing an octogenarian Hamlet .

Perhaps inevitably, there’s a feeling of the star vehicle to this production. In the “sweet creature of bombast” that is this play’s John Falstaff, McKellen has an especially juicy assignment — an outsized character whose appetite for life matches the actor’s own gusto. We’re told that the ample Falstaff hasn’t seen his own knees in years, and when he sits, it looks as if he may never stand up. His mouth, however, seems always in motion, as if chewing food for constant fuel.

He’s also a necessary soul mate to the carousing, drug-using Prince Hal (the excellent Toheeb Jimoh, an Emmy nominee for “ Ted Lasso ”), whose coming-of-age story — becoming, as he puts it, “more myself” — connects these two “Henry IV” plays. But the roustabout Hal’s dawning maturity costs him the companion he once held dear.

“Banish plump Jack, and banish all the world,” says a wary Falstaff in the production’s second half, instructions Hal follows in one of the most ruthless of all Shakespeare’s scenes. And Jimoh — who played a notable Romeo last year at the Almeida — once again shows a fluency with Renaissance language that bodes well for his own Shakespearean future.

Icke hasn’t directed Shakespeare in London since 2017, when Andrew Scott was his Hamlet . Like that show, “Player Kings” uses modern dress, but it forsakes video and hand-held cameras for a comparatively straightforward approach; Icke’s directorial hand is less obvious in this production. Captions appear now and again above the stage to tell us where we are, and Hildegard Bechtler’s brick wall set features curtains pulled across the breadth of the stage to change locations.

Contemporary resonances are inescapable. Watching the party boy Hal of the earlier scenes, whose devil-may-care recklessness will fall away with time, you can’t help but think of Britain’s own Prince Harry, and his onetime reputation as a royal bad boy.

The aging King Henry (an ashen Richard Coyle) is a fretful, anxious figure who might send any child into the contrastingly exuberant embrace of Falstaff, who is a far more pleasing father figure. (Referencing the vilifications of “base news-mongers,” Hal, too, could well be admonishing today’s tabloid press.)

Icke’s adaptation never lets us forget that war is raging in the background. Rebellion, battle and bloodshed are rarely far from view, and there’s a disturbing moment — not to be revealed — in which Falstaff’s capacity for cruelty comes to the fore.

Part 2, as written, is the longer of the plays and can be the more satisfying: more poetic and reflective, less of a rowdy action movie in embryo. That isn’t the case on this occasion, where cuts suggest a desire to speed through to the ending, and the female characters — Doll Tearsheet (Tafline Steen) and Mistress Quickly (Clare Perkins) — seem sidelined in particular.

You come to miss, too, the feral presence of Hotspur (Samuel Edward-Cook), Hal’s rival whose death brings Part 1 to its climax. (The charismatic Edward-Cook doubles as Falstaff’s swaggering sidekick, Pistol.)

These shortfalls are unlikely to matter much to playgoers drawn by the above-the-title star, who has waited a lifetime to play Falstaff after playing two other roles in Part 2 during his student days: Indeed, McKellen’s biography in the playbill cites only his Shakespeare credits — a singularly impressive list.

McKellen revels at every moment in the language that pours luxuriantly forth from Falstaff, whose fondness for verbal embellishment and exaggeration is part of his charm. But you equally feel the neediness that propels Falstaff to savor every experience, carnal ones included.

Why not, then, hand this role over to a life force of the English theater? “You bear your years very well,” Justice Shallow (Robin Soans) says in greeting the wild-haired Falstaff midway through Part 2. His audience would surely drink to that.

Player Kings

Through June 22 at the Noël Coward Theater in London, then touring; playerkingstheplay.co.uk .

Arts and Culture Across Europe

New productions of “Macbeth” and “Hamlet” in Paris follow a French tradition of adapting familiar works . The results are innovative, and sometimes cryptic.

The internet latched on to 16-year-old Felicia Dawkins’ performance as The Unknown at a shambolic Willy Wonka-inspired event . Now she’s heading to a bigger and scarier stage in London.

When activists urged Tate Britain in London to take an offensive artwork off its walls, the institution commissioned Keith Piper  to create a response instead. The result recently went on display.

The new National Holocaust Museum in Amsterdam has been in the works for almost 20 years. It is the first institution to tell the full story  of the persecution of Dutch Jews during World War II.

At a retrospective of John Singer Sargent’s portraits in London, where the American expatriate fled after creating a scandal in Paris, clothes offer both armor and self-expression .

The street artist Frank “Frankey” de Ruwe has been delighting Amsterdam with his whimsical, witty pieces .