(c) We factor out the GCF of the three terms, which is x². 3x4 – 5x3 + x2 = x²(3x²) – x²([x) + x²(1) Write the last term, x?, as x²(1) = x2(Ox2 - 5x + 1) Factor out the GCF, x2. We check by multiplying: x2(3x O) = 3x4 - 5x3 + x². - 5x + 3 Factor: (a) 5f + 45 (b) 16s²t² – 40s³t and (c) y6 – 9y4 – y3 (a) (b) (c)
(c) We factor out the GCF of the three terms, which is x². 3x4 – 5x3 + x2 = x²(3x²) – x²([x) + x²(1) Write the last term, x?, as x²(1) = x2(Ox2 - 5x + 1) Factor out the GCF, x2. We check by multiplying: x2(3x O) = 3x4 - 5x3 + x². - 5x + 3 Factor: (a) 5f + 45 (b) 16s²t² – 40s³t and (c) y6 – 9y4 – y3 (a) (b) (c)
Elementary Algebra
17th Edition
ISBN:9780998625713
Author:Lynn Marecek, MaryAnne Anthony-Smith
Publisher:Lynn Marecek, MaryAnne Anthony-Smith
Chapter6: Polynomials
Section6.1: Add And Subtract Polynomials
Problem 6.29TI: The polynomial 6x2+15xy gives the cost, in dollars, of producing a rectangular container whose top...
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I need help with all of them plz.
![(c) We factor out the GCF of the three terms, which is x².
3x4 – 5x3 + x² = x²(3x²) – x²([ ]x) + x²(1) Write the last term, x², as x²(1)
= x2(Ox2 - 5x + 1)
Factor out the GCF, x².
We check by multiplying: x2(3x² – 5x +
|) = 3x4 – 5x3 + x².
3
Factor: (a) 5f + 45 (b) 16s²t² – 40s³t and (c) y6 – 9yª – y3
(a)
(b)
(c)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9a233757-ff48-478f-9071-7d7d23b994ff%2Fca31cac2-059e-4392-ac48-de9bc73061f1%2Firvkb0p_processed.png&w=3840&q=75)
Transcribed Image Text:(c) We factor out the GCF of the three terms, which is x².
3x4 – 5x3 + x² = x²(3x²) – x²([ ]x) + x²(1) Write the last term, x², as x²(1)
= x2(Ox2 - 5x + 1)
Factor out the GCF, x².
We check by multiplying: x2(3x² – 5x +
|) = 3x4 – 5x3 + x².
3
Factor: (a) 5f + 45 (b) 16s²t² – 40s³t and (c) y6 – 9yª – y3
(a)
(b)
(c)
![Factor: (a) 8m + 24, (b) 35a³b² - 14a?b³, and (c) 3x4 – 5x3 + x2
First, we will determine the GCF of the terms of the polynomial. Then we will write each term of the polynomial as the product of the GCF and one other factor.
We can then use the distributive property to factor out the GCF.
(a) Since the GCF of 8m and 24 is 8, we write 8m and 24 as the product of 8 and one other factor.
8m + 24 = 8•m+ 8•3
(m + 3)
Factor out the GCF, 8.
To check, we multiply: 8(m + 3) = 8 · m + 8 • 3 = 8m + 24. Since we obtain the original polynomial, 8m + 24, the factorization is correct.
Remember to factor out the greatest common factor, not just a common factor. If we factored out 4 in the previous example, we would get
8m + 24 = 4(2m + 6)
However, the terms in red within parentheses have a common factor of 2, indicating that the factoring is not complete.
(b) First, find the GCF of 35a³b² and 14a²b³.
35a3b2 = :7•a•a•a•b•b
The GCF is 7a²b²
14a?b3 =||:7•a•a•b •b•b
%3D
Now, we write 35a b2 and 14a?b3 as the product of the GCF, 7a2b?, and one other factor.
35a³b2 – 14a²b3 = 7a²b² •
– 7a²b² • 2b
а —
= 7a?b?(5a -O b)
7a?b?.
Factor out the GCF,
We check by multiplying: 7a?b?(5a – 2b) = 35a³b² –
a?b³.
%D
(c) We factor out the GCF of the three terms, which is x2.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9a233757-ff48-478f-9071-7d7d23b994ff%2Fca31cac2-059e-4392-ac48-de9bc73061f1%2Fvac535_processed.png&w=3840&q=75)
Transcribed Image Text:Factor: (a) 8m + 24, (b) 35a³b² - 14a?b³, and (c) 3x4 – 5x3 + x2
First, we will determine the GCF of the terms of the polynomial. Then we will write each term of the polynomial as the product of the GCF and one other factor.
We can then use the distributive property to factor out the GCF.
(a) Since the GCF of 8m and 24 is 8, we write 8m and 24 as the product of 8 and one other factor.
8m + 24 = 8•m+ 8•3
(m + 3)
Factor out the GCF, 8.
To check, we multiply: 8(m + 3) = 8 · m + 8 • 3 = 8m + 24. Since we obtain the original polynomial, 8m + 24, the factorization is correct.
Remember to factor out the greatest common factor, not just a common factor. If we factored out 4 in the previous example, we would get
8m + 24 = 4(2m + 6)
However, the terms in red within parentheses have a common factor of 2, indicating that the factoring is not complete.
(b) First, find the GCF of 35a³b² and 14a²b³.
35a3b2 = :7•a•a•a•b•b
The GCF is 7a²b²
14a?b3 =||:7•a•a•b •b•b
%3D
Now, we write 35a b2 and 14a?b3 as the product of the GCF, 7a2b?, and one other factor.
35a³b2 – 14a²b3 = 7a²b² •
– 7a²b² • 2b
а —
= 7a?b?(5a -O b)
7a?b?.
Factor out the GCF,
We check by multiplying: 7a?b?(5a – 2b) = 35a³b² –
a?b³.
%D
(c) We factor out the GCF of the three terms, which is x2.
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