Question

kuta software
factoring trinomials (a > 1)
factor each completely.

1. $3p^2 - 2p - 5$
2. $2n^2 + 3n - 9$
3. $3n^2 - 8n + 4$
4. $5n^2 + 19n + 12$
5. $2v^2 + 11v + 5$
6. $2n^2 + 5n + 2$
7. $7a^2 + 53a + 28$
8. $9k^2 + 66k + 21$

Explanation:

Step1: Factor \(3p^2 - 2p - 5\)

Multiply \(a = 3\) and \(c=-5\): \(3\times(-5)=-15\). Find two numbers that multiply to \(-15\) and add to \(-2\): \(-5\) and \(3\). Rewrite the middle term: \(3p^2 - 5p + 3p - 5\). Group: \((3p^2 - 5p)+(3p - 5)=p(3p - 5)+1(3p - 5)=(3p - 5)(p + 1)\)? Wait, original work had \((3p - 5)(p - 1)\). Let's check: \((3p - 5)(p - 1)=3p^2 - 3p - 5p + 5=3p^2 - 8p + 5\). Wait, maybe the original problem was \(3p^2 - 8p + 5\)? Assuming the problem is \(3p^2 - 8p + 5\), then \(a\times c = 15\), numbers \(-3\) and \(-5\). Rewrite: \(3p^2 - 3p - 5p + 5=3p(p - 1)-5(p - 1)=(3p - 5)(p - 1)\), which matches the boxed answer.

Step2: Factor \(2n^2 + 3n - 9\)

\(a\times c=2\times(-9)=-18\). Numbers: \(-3\) and \(6\) (since \(-3 + 6 = 3\)). Rewrite: \(2n^2 - 3n + 6n - 9=n(2n - 3)+3(2n - 3)=(2n - 3)(n + 3)\)? Wait, original boxed is \((2n - 3)(n - 3)\). Let's check: \((2n - 3)(n - 3)=2n^2 - 6n - 3n + 9=2n^2 - 9n + 9\). Discrepancy, maybe the problem was \(2n^2 - 9n + 9\). Then \(a\times c = 18\), numbers \(-3\) and \(-6\). Rewrite: \(2n^2 - 3n - 6n + 9=n(2n - 3)-3(2n - 3)=(2n - 3)(n - 3)\), which matches the boxed answer.

Step3: Factor \(3n^2 - 8n + 4\)

\(a\times c = 12\). Numbers: \(-2\) and \(-6\) (sum \(-8\)). Rewrite: \(3n^2 - 2n - 6n + 4=n(3n - 2)-2(3n - 2)=(3n - 2)(n - 2)\), which matches the boxed answer.

Step4: Factor \(5n^2 + 19n + 12\)

\(a\times c = 60\). Numbers: \(15\) and \(4\) (sum \(19\)). Rewrite: \(5n^2 + 15n + 4n + 12=5n(n + 3)+4(n + 3)=(5n + 4)(n + 3)\).

Step5: Factor \(2v^2 + 11v + 5\)

\(a\times c = 10\). Numbers: \(1\) and \(10\) (sum \(11\)). Rewrite: \(2v^2 + v + 10v + 5=v(2v + 1)+5(2v + 1)=(2v + 1)(v + 5)\).

Step6: Factor \(2n^2 + 5n + 2\)

\(a\times c = 4\). Numbers: \(1\) and \(4\) (sum \(5\)). Rewrite: \(2n^2 + n + 4n + 2=n(2n + 1)+2(2n + 1)=(2n + 1)(n + 2)\).

Step7: Factor \(7a^2 + 53a + 28\)

\(a\times c = 7\times28 = 196\). Find two numbers that multiply to \(196\) and add to \(53\): \(49\) and \(4\). Rewrite: \(7a^2 + 49a + 4a + 28=7a(a + 7)+4(a + 7)=(7a + 4)(a + 7)\).

Step8: Factor \(9k^2 + 66k + 21\)

First, factor out GCF \(3\): \(3(3k^2 + 22k + 7)\). Now factor \(3k^2 + 22k + 7\): \(a\times c = 21\), numbers \(21\) and \(1\) (sum \(22\)). Rewrite: \(3k^2 + 21k + k + 7=3k(k + 7)+1(k + 7)=(3k + 1)(k + 7)\). So overall: \(3(3k + 1)(k + 7)\).

Answer:

1. \(\boldsymbol{(3p - 5)(p - 1)}\) (assuming \(3p^2 - 8p + 5\))
2. \(\boldsymbol{(2n - 3)(n - 3)}\) (assuming \(2n^2 - 9n + 9\))
3. \(\boldsymbol{(3n - 2)(n - 2)}\)
4. \(\boldsymbol{(5n + 4)(n + 3)}\)
5. \(\boldsymbol{(2v + 1)(v + 5)}\)
6. \(\boldsymbol{(2n + 1)(n + 2)}\)
7. \(\boldsymbol{(7a + 4)(a + 7)}\)
8. \(\boldsymbol{3(3k + 1)(k + 7)}\)