Quadratic expressions

To factorise a quadratic in the form $x^2+bx+c$, we can use factoring by inspection. Start by looking at the expression:

$x^2+4x-21$

The coefficient of $x$ is $b=4$, and the constant term is $c=-21$. We must find integers $p$ and $q$ such that:

1. $p\times q=-21$
2. $p+q=4$

Write down factor pairs of $-21$ and test their sums:

• $-1$ and $21$: Sum $=20$ (Not $4$)
• $1$ and $-21$: Sum $=-20$ (Not $4$)
• $-3$ and $7$: Sum $=4$ (Works!)
• $3$ and $-7$: Sum $=-4$ (Not $4$)

So the numbers we need are $7$ and $-3$.

Now compare with the answer choices.

In Option A, multiplying the constants gives $(-7)\times (-3)=+21$, but we need $-21$, so it cannot be correct.

In Option B, the middle term would come from $21-1=20$, giving $+20x$ rather than $+4x$, so it is also not correct.

Consider Option D. Expanding $(x-7)(x+3)$ gives a middle term of $-7x+3x=-4x$. Our expression needs $+4x$, so Option D is wrong.

Because the pair is $+7$ and $-3$, we write the factorised form as:

$(x+7)(x-3)$

This matches the remaining option and expands back to $x^2+4x-21$.

Another approach is to expand each proposed factorisation using the FOIL method (First, Outside, Inside, Last). We are looking for the one that expands to:

$x^2+4x-21$

Expand Option A: $(x-7)(x-3)$

$\begin{array}{rl}(x-7)(x-3)& =x(x)+x(-3)-7(x)-7(-3)\\ & =x^2-3x-7x+21\\ & =x^2-10x+21\end{array}$

This becomes $x^2-10x+21$, which doesn’t match the $+4x$ middle term or the $-21$ constant in the original expression.

Expand Option B: $(x+21)(x-1)$

$\begin{array}{rl}(x+21)(x-1)& =x(x)+x(-1)+21(x)+21(-1)\\ & =x^2-x+21x-21\\ & =x^2+20x-21\end{array}$

The constant term is $-21$, but the middle term is $+20x$, not $+4x$, so this option is incorrect.

Expand Option D: $(x-7)(x+3)$

$\begin{array}{rl}(x-7)(x+3)& =x(x)+x(3)-7(x)-7(3)\\ & =x^2+3x-7x-21\\ & =x^2-4x-21\end{array}$

This has the correct constant term, but the middle term is $-4x$ instead of $+4x$, so it still doesn’t match.

Expand Option C: $(x+7)(x-3)$

$\begin{array}{rl}(x+7)(x-3)& =x(x)+x(-3)+7(x)+7(-3)\\ & =x^2-3x+7x-21\\ & =x^2+4x-21\end{array}$

This expansion is exactly the same as the original expression. This topic is covered in Functions and Quadratic Relationships in the MYP curriculum.

To factorise $x^2+4x-21$ using the quadratic formula, begin by setting it equal to zero:

$x^2+4x-21=0$

Match this to $a{x}^2+bx+c=0$ to identify:

• $a=1$
• $b=4$
• $c=-21$

Use the quadratic formula from your IB Mathematics data booklet:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $a=1$, $b=4$, and $c=-21$:

$x=\frac{-4\pm \sqrt{4^2-4(1)(-21)}}{2(1)}$

Compute the discriminant: $4^2-4(1)(-21)=16+84=100$

Since $\sqrt{100}=10$, the two solutions are:

$\begin{array}{rl}{x}_1& =\frac{-4+10}{2}=\frac{6}{2}=3\\ x_2& =\frac{-4-10}{2}=\frac{-14}{2}=-7\end{array}$

By the factor theorem, if $x=k$ is a root, then $(x-k)$ is a factor.

• If $x=3$, then the factor is $(x-3)$.
• If $x=-7$, then the factor is $(x-(-7))=(x+7)$.

So the factorised form is: $(x+7)(x-3)$

(You can swap the factors and it still represents the same product.)

Our result is $(x+7)(x-3)$, so:

• Option A is wrong because it uses the incorrect sign with $7$.
• Option B is wrong because it uses $21$ and $1$ instead of $7$ and $3$.
• Option D is wrong because the signs are arranged to give the wrong middle term.

$(x+7)(x-3)$ corresponds to Option C.

In mathematics, an identity is an equality that stays true no matter what values are put in for the variable(s). For instance, $2(x+3)=2x+6$ is an identity because it works for every possible value of $x$.

A typical equation (often called a conditional equation) is an equality that only works for certain value(s). For example, $2x+1=7$ is true only when $x=3$. Substituting a different value of $x$ makes the statement false.

Option A says identities must always be written equal to zero. Although you can rearrange an identity so one side is zero (like $x-x=0$), that is not what defines an identity. For example, sin^2(x) + cos^2(x) = 1 is an identity and it is not set to zero, so Option A is not correct.

Option C says “identity” and “equation” are interchangeable. But the definitions show they behave differently: identities are true for all inputs, while equations are true only for their solution set. So they are not synonyms.

Option D claims the opposite of the formal meanings: it says equations are always true and identities are only sometimes true. That reverses the definitions we recalled, so Option D must be wrong.

Using the definitions about which values make the statement true, an identity holds for all values in the domain, while a conditional equation holds only for specific solution value(s).

To tell an identity apart from an equation, think about what happens when you substitute different numbers.

• Identity Example: Consider $2(x+1)=2x+2$. Whatever value you choose for $x$, both sides match.
• Equation Example: Consider $x+2=5$. This only works if $x=3$. If we try $x=1$, it becomes false because $1+2\ne 5$.

We can now use these ideas to remove the wrong choices.

Option C says the two terms mean the same thing.

But in algebra they refer to different kinds of truth: some equalities are true for every input (identities), and others are true only when the variable takes specific solution values (equations). So Option C cannot be right.

Option A claims an identity must be written with a zero on one side.

Check this well-known trig identity: ${\sin }^2(\theta )+{\cos }^2(\theta )=1$

It is true for any angle $\theta$, yet it equals $1$, not $0$. So identities do not have to be “set to zero,” making Option A incorrect.

Option D says an equation is true for all values, and an identity is only true for some values.

That contradicts the definitions. For instance, the equation $3x=12$ is only true when $x=4$, not for every value (certainly not for $x=0$ or $x=1$). So Option D is eliminated.

With the other options removed, Option B remains and it matches the definitions:

• Identity: Always true for any value of the variable(s) (e.g., $x+x=2x$).
• Equation: True only for specific value(s) that solve it (e.g., $2x=10$).