Question

worksheet: factoring quadratics - is (x - a) a factor? name:leandro simbolio date: 8 - 9 - 25 instructions: for each quadratic function below, determine whether the given binomial is a factor. show your work and explain your reasoning using substitution. example: given: f(x)=x² - 5x + 6 check if (x - 2) is a factor. f(2)=2² - 5(2)+6 = 0→so, (x - 2) is a factor. questions. solve at the back page. 1. f(x)=x² - 7x + 10—is (x - 2) a factor? yes 2. f(x)=x² + 3x - 10—is (x + 5) a factor? yes 3. f(x)=x² - 4x - 5—is (x + 1) a factor? yes 4. f(x)=x² - 6x + 9—is (x - 3) a factor? 5. f(x)=x² + x - 12—is (x + 3) a factor? 6. f(x)=2x² - 5x + 2—is (x - 2) a factor? 7. f(x)=3x² - 2x - 1—is (x + 1) a factor? 8. f(x)=x² + 2x + 1—is (x + 1) a factor? 9. f(x)=x² - 9—is (x + 3) a factor? 10.f(x)=x² - x - 6—is (x - 3) a factor? 11.f(x)=x² - 10x + 25—is (x - 5) a factor? 12.f(x)=x² + 4x + 3—is (x + 3) a factor? 13.f(x)=x² - x - 2—is (x - 2) a factor? 14.f(x)=4x² - 4x - 3—is (x - 1) a factor? 15.f(x)=x² + 5x + 6—is (x + 2) a factor? bonus challenge: f(x)=x² + kx + 6 find the value(s) of k such that (x + 2) is a factor of f(x). let f(x)=x² + kx + 12. find the value(s) of k such that (x + 3) is a factor of f(x).

Explanation:

Step1: Recall factor - theorem

If \((x - a)\) is a factor of \(f(x)\), then \(f(a)=0\).

Step2: Solve question 4

Given \(f(x)=x^{2}-6x + 9\) and we want to check if \((x - 3)\) is a factor. Substitute \(x = 3\) into \(f(x)\):
\[f(3)=3^{2}-6\times3 + 9=9-18 + 9=0\]
So \((x - 3)\) is a factor.

Step3: Solve question 5

Given \(f(x)=x^{2}+x - 12\) and we want to check if \((x + 3)\) is a factor. Substitute \(x=-3\) into \(f(x)\):
\[f(-3)=(-3)^{2}+(-3)-12=9 - 3-12=-6
eq0\]
So \((x + 3)\) is not a factor.

Step4: Solve question 6

Given \(f(x)=2x^{2}-5x + 2\) and we want to check if \((x - 2)\) is a factor. Substitute \(x = 2\) into \(f(x)\):
\[f(2)=2\times2^{2}-5\times2 + 2=2\times4-10 + 2=8-10 + 2=0\]
So \((x - 2)\) is a factor.

Step5: Solve question 7

Given \(f(x)=3x^{2}-2x - 1\) and we want to check if \((x + 1)\) is a factor. Substitute \(x=-1\) into \(f(x)\):
\[f(-1)=3\times(-1)^{2}-2\times(-1)-1=3 + 2-1=4
eq0\]
So \((x + 1)\) is not a factor.

Step6: Solve question 8

Given \(f(x)=x^{2}+2x + 1\) and we want to check if \((x + 1)\) is a factor. Substitute \(x=-1\) into \(f(x)\):
\[f(-1)=(-1)^{2}+2\times(-1)+1=1-2 + 1=0\]
So \((x + 1)\) is a factor.

Step7: Solve question 9

Given \(f(x)=x^{2}-9\) and we want to check if \((x + 3)\) is a factor. Substitute \(x=-3\) into \(f(x)\):
\[f(-3)=(-3)^{2}-9=9-9=0\]
So \((x + 3)\) is a factor.

Step8: Solve question 10

Given \(f(x)=x^{2}-x - 6\) and we want to check if \((x - 3)\) is a factor. Substitute \(x = 3\) into \(f(x)\):
\[f(3)=3^{2}-3-6=9-3-6=0\]
So \((x - 3)\) is a factor.

Step9: Solve question 11

Given \(f(x)=x^{2}-10x + 25\) and we want to check if \((x - 5)\) is a factor. Substitute \(x = 5\) into \(f(x)\):
\[f(5)=5^{2}-10\times5 + 25=25-50 + 25=0\]
So \((x - 5)\) is a factor.

Step10: Solve question 12

Given \(f(x)=x^{2}+4x + 3\) and we want to check if \((x + 3)\) is a factor. Substitute \(x=-3\) into \(f(x)\):
\[f(-3)=(-3)^{2}+4\times(-3)+3=9-12 + 3=0\]
So \((x + 3)\) is a factor.

Step11: Solve question 13

Given \(f(x)=x^{2}-x - 2\) and we want to check if \((x - 2)\) is a factor. Substitute \(x = 2\) into \(f(x)\):
\[f(2)=2^{2}-2-2=4-2-2=0\]
So \((x - 2)\) is a factor.

Step12: Solve question 14

Given \(f(x)=4x^{2}-4x - 3\) and we want to check if \((x - 1)\) is a factor. Substitute \(x = 1\) into \(f(x)\):
\[f(1)=4\times1^{2}-4\times1-3=4-4-3=-3
eq0\]
So \((x - 1)\) is not a factor.

Step13: Solve question 15

Given \(f(x)=x^{2}+5x + 6\) and we want to check if \((x + 2)\) is a factor. Substitute \(x=-2\) into \(f(x)\):
\[f(-2)=(-2)^{2}+5\times(-2)+6=4-10 + 6=0\]
So \((x + 2)\) is a factor.

Step14: Solve bonus challenge 1

Given \(f(x)=x^{2}+kx + 6\) and \((x + 2)\) is a factor. Then \(f(-2)=0\).
\[f(-2)=(-2)^{2}+k\times(-2)+6=0\]
\[4-2k + 6=0\]
\[10-2k=0\]
\[2k=10\]
\[k = 5\]

Step15: Solve bonus challenge 2

Given \(f(x)=x^{2}+kx + 12\) and \((x + 3)\) is a factor. Then \(f(-3)=0\).
\[f(-3)=(-3)^{2}+k\times(-3)+12=0\]
\[9-3k + 12=0\]
\[21-3k=0\]
\[3k=21\]
\[k = 7\]

Answer:

1. Yes
2. No
3. Yes
4. No
5. Yes
6. Yes
7. Yes
8. Yes
9. Yes
10. Yes
11. No
12. Yes

Bonus 1: \(k = 5\)
Bonus 2: \(k = 7\)