Leaving Cert Mathematics Functions: (x + 4) is a factor of $f(x) = x

(x + 4) is a factor of $f(x) = x^3 + qx^2 - 22x + 56$, where $x \in \mathbb{R}$ and $q \in \mathbb{Z}$ - Leaving Cert Mathematics - Question b - 2021

Question b

$(x-+-4)-is-a-factor-of-$f(x)-=-x^3-+-qx^2---22x-+-56$,-where-$x-\in-\mathbb{R}$-and-$q-\in-\mathbb{Z}$-Leaving Cert Mathematics-Question b-2021.png$

(x + 4) is a factor of $f(x) = x^3 + qx^2 - 22x + 56$, where $x \in \mathbb{R}$ and $q \in \mathbb{Z}$. Show that $q = -5$, and find the three roots of $f(x)$. Show... show full transcript

Worked Solution & Example Answer:(x + 4) is a factor of $f(x) = x^3 + qx^2 - 22x + 56$, where $x \in \mathbb{R}$ and $q \in \mathbb{Z}$ - Leaving Cert Mathematics - Question b - 2021

Step 1

Show that $q = -5$

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Answer

To confirm that $(x + 4)$ is a factor of the polynomial $f(x)$ , we need to perform polynomial long division. First, we substitute $x = -4$ into the function:

$f(-4) = (-4)^3 + q(-4)^2 - 22(-4) + 56$

Calculating the known values gives us:

$f(-4) = -64 + 16q + 88 + 56$

Simplifying this, we get:

$f(-4) = 16q + 80$

Setting this equal to zero for $(x + 4)$ to be a factor, we have:

$16q + 80 = 0 \Rightarrow 16q = -80 \Rightarrow q = -5.$

Step 2

Find the three roots of $f(x)$

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Answer

Now that we have $q = -5$ , we can write the function as:

$f(x) = x^3 - 5x^2 - 22x + 56.$

Next, we perform polynomial long division by dividing $f(x)$ by $(x + 4)$ :

Divide the leading term: $x^3 ÷ x = x^2$
Multiply $(x + 4)$ by $x^2$ : $x^3 + 4x^2$
Subtract from $f(x)$ : $f(x) - (x^3 + 4x^2) = -5x^2 - 4x^2 - 22x + 56 = -9x^2 - 22x + 56.$
Divide the leading term: $-9x^2 ÷ x = -9x$ .
Multiply $(x + 4)$ by $-9x$ : $-9x^2 - 36x$ .
Subtract: $-9x^2 - 22x + 56 - (-9x^2 - 36x) = 14x + 56.$
Divide the leading term: $14x ÷ x = 14$ .
Multiply $(x + 4)$ by $14$ : $14x + 56$ .
Subtract: $14x + 56 - (14x + 56) = 0.$

Thus, we have:

$f(x) = (x + 4)(x^2 - 9x + 14).$

Next, we find the roots of the quadratic $x^2 - 9x + 14$ . Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{9 \pm \sqrt{(-9)^2 - 4(1)(14)}}{2(1)} = \frac{9 \pm \sqrt{81 - 56}}{2} = \frac{9 \pm \sqrt{25}}{2} = \frac{9 \pm 5}{2}.$

Calculating the roots:

$x = \frac{14}{2} = 7.$
$x = \frac{4}{2} = 2.$
The three roots of $f(x)$ are therefore $(4, 2, 7).$

(x + 4) is a factor of $f(x) = x^3 + qx^2 - 22x + 56$, where $x \in \mathbb{R}$ and $q \in \mathbb{Z}$ - Leaving Cert Mathematics - Question b - 2021

Worked Solution & Example Answer:(x + 4) is a factor of $f(x) = x^3 + qx^2 - 22x + 56$, where $x \in \mathbb{R}$ and $q \in \mathbb{Z}$ - Leaving Cert Mathematics - Question b - 2021

Show that $q = -5$

Find the three roots of $f(x)$

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