Math

QuestionFactor these expressions completely:
1. 5z2+38z+215 z^{2}+38 z+21
2. 16w3+20w2+6w16 w^{3}+20 w^{2}+6 w
3. 20x424x332x220 x^{4}-24 x^{3}-32 x^{2}

Studdy Solution

STEP 1

Assumptions1. The given expression is 5z+38z+215z^ +38z +21 . We need to factor this expression completely

STEP 2

The given expression is a quadratic equation of the form ax2+bx+cax^2 + bx + c. We can factorize it using the method of splitting the middle term.

STEP 3

To split the middle term, we need to find two numbers such that their product is equal to the product of aa and cc, and their sum is equal to bb.In this case, a=5a =5, b=38b =38, and c=21c =21. So, we need to find two numbers whose product is 5×21=1055 \times21 =105 and whose sum is 3838.

STEP 4

The numbers that satisfy these conditions are 3535 and 33, because 35×3=10535 \times3 =105 and 35+3=3835 +3 =38.

STEP 5

Now, we can split the middle term of the quadratic equation using these two numbers. This gives us5z2+35z+3z+215z^2 +35z +3z +21

STEP 6

Next, we group the terms to factor by grouping5z2+35z+3z+21=5z(z+)+3(z+)5z^2 +35z +3z +21 =5z(z +) +3(z +)

STEP 7

Notice that (z+7)(z +7) is a common factor. We can factor that out, giving us5z(z+7)+3(z+7)=(5z+3)(z+7)5z(z +7) +3(z +7) = (5z +3)(z +7)So, the completely factored form of the given expression is (5z+3)(z+7)(5z +3)(z +7).

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