FACTOR USING QUADRATIC PATTERN

Factor the following 4th degree polynomials using quadratic pattern.

Example 1 :

x⁴ - 7x² - 18

Solution :

  =  x⁴ - 7x² - 18

  =  (x²)² - 7x² - 18

Let x² = t

  =  t² - 7t - 18

Step 1 :

By multiplying the coefficient of x² by the constant term -18, we get -18.

Step 2 :

Now we need to spit this -18 as two parts, and the product of those parts must be equal to -18 and simplified value must be equal to the middle term (-7).

Since the middle and last terms are negative, the factors will be in the combination of positive and negative.

-18  =  -9 ⋅ 2 and -9 + 2 =  -7

Step 3 :

Grouping into linear factors

=  t² - 9t + 2t - 18

=  t (t - 9) + 2(t - 9)

=  (t + 2) (t - 9)

Step 4 :

Replacing "t" by x²,

=  (x² + 2) (x² - 9)

Hence the factors are (x² + 2) (x² - 9).

Example 2 :

x⁴ - 16x² + 63

Solution :

  =  x⁴ - 16x² + 63

  =  (x²)² - 16x² + 63

Let x² = t

  =  t²  - 16t + 63

Step 1 :

By multiplying the coefficient of x² by the constant term 63, we get 63.

Step 2 :

Now we need to spit this 63 as two parts, and the product of those parts must be equal to 63 and simplified value must be equal to the middle term (-16).

Since the middle is negative, both factors will have negative sign.

63  =  -9 ⋅ (-7) and -9 + (-7) =  -16

Step 3 :

Grouping into linear factors

=  t² - 9t - 7t + 63

=  t (t - 9) - 7(t - 9)

=  (t - 9) (t - 7)

Step 4 :

Replacing "t" by x²,

=  (x² - 9) (x² - 7)

Hence the factors are (x² - 9) (x² - 7).

Example 3 :

7x⁴ - 45x² - 28

Solution :

  =  x⁴ - 45x² - 28

  =  7(x²)² - 45x² - 28

Let x² = t

  =  7t²  - 45t - 28

Step 1 :

By multiplying the coefficient of x² (7) by the constant term 28, we get -196.

Step 2 :

Now we need to spit this -196 as two parts, and the product of those parts must be equal to -196 and simplified value must be equal to the middle term (-45).

Since the middle and last terms are negative, the factors will be in the combination of positive and negative.

-196  =  -49 ⋅ 4 and -49 + 4 =  -45

Step 3 :

Grouping into linear factors

  =  7t²  - 45t - 28

  =  7t²  - 49t + 4t - 28

=  7t (t - 7) + 4(t - 7)

=  (7t + 4) (t - 7)

Step 4 :

Replacing "t" by x²,

=  (7x² + 4) (x² - 7)

Hence the factors are (7x² + 4) (x² - 7).

Example 4 :

5x⁴ - x² - 18

Solution :

  =  5x⁴ - x² - 18

  =  5(x²)² - x² - 18

Let x² = t

  =  5t²  - t - 18

Step 1 :

By multiplying the coefficient of x² (5) by the constant term -18, we get -90.

Step 2 :

Now we need to spit this -90 as two parts, and the product of those parts must be equal to -90 and simplified value must be equal to the middle term (-1).

Since the middle and last terms are negative, the factors will be in the combination of positive and negative.

-90  =  -10 ⋅ 9 and -10 + 9 =  -1

Step 3 :

Grouping into linear factors

  =  5t²  - t - 18

  =  5t²  - 10t  + 9t - 18

=  5t (t - 2) + 9(t - 2)

=  (5t + 9) (t - 2)

Step 4 :

Replacing "t" by x²,

=  (5x² + 9) (x² - 2)

Hence the factors are  (5x² + 9) (x² - 2).

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