Exercise 46 Page 301

To solve the given equation by factoring, we will start by writing all the terms on the left-hand side.

27 = - x^{4} - 12 x^{2}

AddEqn

$LHS + x^{4} = RHS + x^{4}$

x^{4} + 27 = - 12 x^{2}

AddEqn

$LHS + 12 x^{2} = RHS + 12 x^{2}$

x^{4} + 12 x^{2} + 27 = 0

Now we need to define another variable. If we let

z = x^{2},

we can rewrite the equation in terms of the

z -

variable.

x^{4} + 12 x^{2} + 27 = 0 \Leftrightarrow z^{2} + 12 z + 27 = 0

Note that the above equation in terms of

z

is a quadratic equation. Therefore, we can solve it using the Quadratic Formula.

a z^{2} + b z + c = 0 ⇕ z = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Let's identify

a,

b,

and

c .

1 z^{2} + 12 z + 27 = 0

We see that

a = 1,

b = 12,

and

c = 27 .

Let's substitute these values into the formula and solve for

z .

z = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

z = \frac{- 1 2 \pm 1 2 ^{2} - 4 ( 1 ) ( 2 7 )}{2 ( 1 )}

▼

Solve for

z

CalcPow

Calculate power

z = \frac{- 1 2 \pm 1 4 4 - 4 ( 1 ) ( 2 7 )}{2 ( 1 )}

IdPropMult

Identity Property of Multiplication

z = \frac{- 1 2 \pm 1 4 4 - 4 ( 2 7 )}{2}

Multiply

z = \frac{- 1 2 \pm 1 4 4 - 1 0 8}{2}

SubTerm

Subtract term

z = \frac{- 1 2 \pm 3 6}{2}

CalcRoot

Calculate root

z = \frac{- 1 2 \pm 6}{2}

FactorOut

Factor out $2$

z = \frac{2 ( - 6 \pm 3 )}{2}

CancelCommonFac

Cancel out common factors

z = - 6 \pm 3

The solutions for this equation are

z = - 6 \pm 3 .

Let's separate them by using the positive and negative signs.

$z = - 6 \pm 3$
$z = - 6 + 3$	$z = - 6 - 3$
$z = - 3$	$z = - 9$

We found that the solutions for

z^{2} + 12 z + 27 = 0

are

z = - 3

and

z = - 9 .

This means that

x^{2} = - 3

and

x^{2} = - 9 .

Let's solve these equations!

x^{2} = - 3 x^{2} = - 9 (I) (II)

$(I), (II):$ $LHS = RHS$

x = \pm - 3 x = \pm - 9

▼

Simplify

$(I), (II):$ Split into factors

x = \pm (- 1) \cdot 3 x = \pm 9 \cdot (- 1)

$(I), (II):$ $a \cdot b = a \cdot b$

x = \pm - 1 \cdot 3 x = \pm 9 \cdot - 1

CalcRoot

$(II):$ Calculate root

x = \pm - 1 \cdot 3 x = \pm 3 - 1

$(I), (II):$ $- 1 = i$

x = \pm i 3 x = \pm 3 i

We found in total four solutions for

x .

x = i 3, x = - i 3, x = 3 i, x = - 3 i

Alternative Solution

Solving

x^{4} + 12 x^{2} + 27 = 0

for

x

Having rewritten the equation so that all the terms on the left-hand side, we can also solve the equation by factoring.

x^{4} + 12 x^{2} + 27 = 0

WriteSum

Write as a sum

x^{4} + 3 x^{2} + 9 x^{2} + 27 = 0

▼

Factor out

x^{2} & 9

FactorOut

Factor out $x^{2}$

x^{2} (x^{2} + 3) + 9 x^{2} + 27 = 0

FactorOut

Factor out $9$

x^{2} (x^{2} + 3) + 9 (x^{2} + 3) = 0

FactorOut

Factor out $(x^{2} + 3)$

(x^{2} + 3) (x^{2} + 9) = 0

To solve for

x

we will again use Zero Product Property.

(x^{2} + 3) (x^{2} + 9) = 0

ZeroProdProp

Use the Zero Product Property

x^{2} = - 3 x^{2} = - 9 (I) (II)

$(I), (II):$ $LHS = RHS$

x = \pm - 3 x = \pm - 9

▼

Simplify

$(I), (II):$ Split into factors

x = \pm (- 1) \cdot 3 x = \pm 9 \cdot (- 1)

$(I), (II):$ $a \cdot b = a \cdot b$

x = \pm - 1 \cdot 3 x = \pm 9 \cdot - 1

CalcRoot

$(II):$ Calculate root

x = \pm - 1 \cdot 3 x = \pm 3 - 1

$(I), (II):$ $- 1 = i$

x = \pm i 3 x = \pm 3 i

We also found the same four solutions for

x .

x = i 3, x = - i 3, x = 3 i, x = - 3 i

Hints

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Alternative Solution