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Math  /  Algebra

Questionh(x)=6x3+19x2−21x−4h(x)=6 x^{3}+19 x^{2}-21 x-4 : Use polynomial division and the quadratic formula, if necessary,

Studdy Solution

STEP 1

What is this asking? Find all the values of xx that make the polynomial h(x)h(x) equal to zero. Watch out! Polynomial division can be tricky, so double-check each step!
Also, remember the quadratic formula has a plus-or-minus, so we might get two answers from that part.

STEP 2

1. Find a root
2. Polynomial Division
3. Quadratic Formula

STEP 3

Let's **test some small values** of xx to see if any of them make h(x)h(x) equal to zero.
Let's try x=2x = 2:
h(2)=6⋅(2)3+19⋅(2)2−21⋅(2)−4 h(2) = 6 \cdot (2)^3 + 19 \cdot (2)^2 - 21 \cdot (2) - 4 h(2)=6⋅8+19⋅4−42−4 h(2) = 6 \cdot 8 + 19 \cdot 4 - 42 - 4 h(2)=48+76−42−4 h(2) = 48 + 76 - 42 - 4 h(2)=78 h(2) = 78

STEP 4

Hmm, not zero.
Let's try x=−4x = -4:
h(−4)=6⋅(−4)3+19⋅(−4)2−21⋅(−4)−4 h(-4) = 6 \cdot (-4)^3 + 19 \cdot (-4)^2 - 21 \cdot (-4) - 4 h(−4)=6⋅(−64)+19⋅16+84−4 h(-4) = 6 \cdot (-64) + 19 \cdot 16 + 84 - 4 h(−4)=−384+304+84−4 h(-4) = -384 + 304 + 84 - 4 h(−4)=0 h(-4) = 0 Woohoo! We found one root: x=−4x = \mathbf{-4}!

STEP 5

Since x=−4x = -4 is a root, we know (x+4)(x+4) must be a factor of h(x)h(x).
Let's use polynomial division to find the other factor.
We're dividing 6x3+19x2−21x−46x^3 + 19x^2 - 21x - 4 by (x+4)(x+4).

STEP 6

After performing the polynomial division, we get 6x2−5x−16x^2 - 5x - 1.
So, we can rewrite h(x)h(x) as:
h(x)=(x+4)(6x2−5x−1) h(x) = (x+4)(6x^2 - 5x - 1)

STEP 7

Now, we need to find the roots of the quadratic 6x2−5x−16x^2 - 5x - 1.
Let's use the **quadratic formula**!
x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} In our case, a=6a = \mathbf{6}, b=−5b = \mathbf{-5}, and c=−1c = \mathbf{-1}.

STEP 8

**Plugging in the values**:
x=−(−5)±(−5)2−4⋅6⋅(−1)2⋅6 x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 6 \cdot (-1)}}{2 \cdot 6} x=5±25+2412 x = \frac{5 \pm \sqrt{25 + 24}}{12} x=5±4912 x = \frac{5 \pm \sqrt{49}}{12} x=5±712 x = \frac{5 \pm 7}{12}

STEP 9

This gives us two possible solutions:
x=5+712=1212=1 x = \frac{5 + 7}{12} = \frac{12}{12} = 1 x=5−712=−212=−16 x = \frac{5 - 7}{12} = \frac{-2}{12} = -\frac{1}{6}

STEP 10

The values of xx that make h(x)=0h(x) = 0 are x=−4x = \mathbf{-4}, x=1x = \mathbf{1}, and x=−1/6x = \mathbf{-1/6}.

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