<span>Equation at the end of step 1 :</span><span> ((x4) - (2•5x2)) + 9 = 0 </span><span>Step 2 :</span>Trying to factor by splitting the middle term
<span> 2.1 </span> Factoring <span> x4-10x2+9</span>
The first term is, <span> <span>x4</span> </span> its coefficient is <span> 1 </span>.
The middle term is, <span> <span>-10x2</span> </span> its coefficient is <span> -10 </span>.
The last term, "the constant", is <span> +9 </span>
Step-1 : Multiply the coefficient of the first term by the constant <span> <span> 1</span> • 9 = 9</span>
Step-2 : Find two factors of 9 whose sum equals the coefficient of the middle term, which is <span> -10 </span>.
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -9 and -1
<span>x4 - 9x2</span> - <span>1x2 - 9</span>
Step-4 : Add up the first 2 terms, pulling out like factors :
<span>x2 • (x2-9)</span>
Add up the last 2 terms, pulling out common factors :
1 • <span>(x2-9)</span>
Step-5 : Add up the four terms of step 4 :
<span>(x2-1) • (x2-9)</span>
Which is the desired factorization
<span> 2.2 </span> Factoring: <span> x2-1</span>
Theory : A difference of two perfect squares, <span> A2 - B2 </span>can be factored into <span> (A+B) • (A-B)
</span>Proof :<span> (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 <span>- AB + AB </span>- B2 =
<span> A2 - B2</span>
</span>Note : <span> <span>AB = BA </span></span>is the commutative property of multiplication.
Note : <span> <span>- AB + AB </span></span>equals zero and is therefore eliminated from the expression.
Check : 1 is the square of 1
Check : <span> x2 </span>is the square of <span> x1 </span>
Factorization is : (x + 1) • (x - 1)
<span> 2.3 </span> Factoring: <span> x2 - 9</span>
Check : 9 is the square of 3
Check : <span> x2 </span>is the square of <span> x1 </span>
Factorization is : (x + 3) • (x - 3)
<span> 3.1 </span> A product of several terms equals zero.<span>
</span>When a product of two or more terms equals zero, then at least one of the terms must be zero.<span>
</span>We shall now solve each term = 0 separately<span>
</span>In other words, we are going to solve as many equations as there are terms in the product<span>
</span>Any solution of term = 0 solves product = 0 as well.
<span> 3.2 </span> Solve : x+1 = 0<span>
</span>Subtract 1 from both sides of the equation :<span>
</span> x = -1
<span> 3.3 </span> Solve : x-1 = 0<span>
</span>Add 1 to both sides of the equation :<span>
</span> x = 1
<span> 3.4 </span> Solve : x+3 = 0<span>
</span>Subtract 3 from both sides of the equation :<span>
</span> x = -3
<span> 3.5 </span> Solve : x-3 = 0<span>
</span>Add 3 to both sides of the equation :<span>
</span> x = 3
Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula
<span>Solving a Single Variable Equation : </span>Equations which are reducible to quadratic :<span> 4.1 </span> Solve <span> x4-10x2+9 = 0</span>
This equation is reducible to quadratic. What this means is that using a new variable, we can rewrite this equation as a quadratic equation Using w , such that <span> w = x2</span> transforms the equation into :
<span> w2-10w+9 = 0</span>
Solving this new equation using the quadratic formula we get two real solutions :
9.0000 or 1.0000
Now that we know the value(s) of <span> w</span> , we can calculate <span> x</span> since <span> x</span> <span> is </span><span> √<span> w </span></span>
Doing just this we discover that the solutions of
<span> x4-10x2+9 = 0</span>
are either :
x =√<span> 9.000 </span>= 3.00000 or :
x =√<span> 9.000 </span>= -3.00000 or :
x =√<span> 1.000 </span>= 1.00000 or :
x =√<span> 1.000 </span>= -1.00000
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