Question

factor completely

1. (4a^{4}b^{3}-12a^{3}b^{2}+28a^{2}bc)
2. (6x^{3}y^{2}-12x^{2}y + 18xy)
3. (12ax + 4bx-9ay-3by)
4. (5x^{2}+35x + 3xy+21y)
5. (4x^{2}+17x-15)
6. (12xy^{2}-2xy - 4x)
7. (27a^{3}+125b^{3})
8. (64x^{6}-y^{6})

Explanation:

Step1: Find GCF for 22.

For the expression $4a^{4}b^{3}-12a^{3}b^{2}+28a^{2}bc$, the GCF of the coefficients 4, - 12, 28 is 4, and the GCF of the variables is $a^{2}b$. So we factor out $4a^{2}b$:
$4a^{2}b(a^{2}b^{2}-3ab + 7c)$

Step2: Find GCF for 23.

For the expression $6x^{3}y^{2}-12x^{2}y + 18xy$, the GCF of the coefficients 6, - 12, 18 is 6, and the GCF of the variables is $xy$. Factoring out $6xy$ gives:
$6xy(x^{2}y-2x + 3)$

Step3: Group - factor for 24.

For $12ax+4bx - 9ay-3by$, group the terms: $(12ax + 4bx)-(9ay + 3by)$. Factor out the GCF from each group: $4x(3a + b)-3y(3a + b)=(3a + b)(4x-3y)$

Step4: Group - factor for 25.

For $5x^{2}+35x + 3xy+21y$, group the terms: $(5x^{2}+35x)+(3xy + 21y)$. Factor out the GCF from each group: $5x(x + 7)+3y(x + 7)=(x + 7)(5x+3y)$

Step5: Factor quadratic for 26.

For $4x^{2}+17x-15$, we need to find two numbers that multiply to $4\times(-15)=-60$ and add up to 17. The numbers are 20 and - 3. Rewrite the middle - term: $4x^{2}+20x-3x - 15$. Group and factor: $4x(x + 5)-3(x + 5)=(x + 5)(4x-3)$

Step6: Find GCF for 27.

For $12xy^{2}-2xy-4x$, the GCF of the coefficients 12, - 2, - 4 is 2, and the GCF of the variables is $x$. Factoring out $2x$ gives: $2x(6y^{2}-y - 2)$. Then factor the quadratic $6y^{2}-y - 2=(2y + 1)(3y-2)$, so the full factorization is $2x(2y + 1)(3y-2)$

Step7: Use sum of cubes for 28.

For $27a^{3}+125b^{3}$, since $27a^{3}=(3a)^{3}$ and $125b^{3}=(5b)^{3}$, using the sum - of - cubes formula $a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})$, we have $(3a + 5b)(9a^{2}-15ab + 25b^{2})$

Step8: Use difference of squares and difference/sum of cubes for 29.

For $64x^{6}-y^{6}=(8x^{3})^{2}-(y^{3})^{2}$. First, use the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$: $(8x^{3}+y^{3})(8x^{3}-y^{3})$. Then use the sum - of - cubes formula $a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})$ for $8x^{3}+y^{3}=(2x + y)(4x^{2}-2xy + y^{2})$ and the difference - of - cubes formula $a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})$ for $8x^{3}-y^{3}=(2x - y)(4x^{2}+2xy + y^{2})$. So $64x^{6}-y^{6}=(2x + y)(2x - y)(4x^{2}-2xy + y^{2})(4x^{2}+2xy + y^{2})$

Answer:

1. $4a^{2}b(a^{2}b^{2}-3ab + 7c)$
2. $6xy(x^{2}y-2x + 3)$
3. $(3a + b)(4x-3y)$
4. $(x + 7)(5x+3y)$
5. $(x + 5)(4x-3)$
6. $2x(2y + 1)(3y-2)$
7. $(3a + 5b)(9a^{2}-15ab + 25b^{2})$
8. $(2x + y)(2x - y)(4x^{2}-2xy + y^{2})(4x^{2}+2xy + y^{2})$