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Algebra 1 : Factoring Polynomials

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Example Questions

Example Question #1 : How To Factor A Polynomial

Factor the following:

x^2-7x+6

Possible Answers:

(x-6)(x-1)

None of the available answers

(x-6)(x+1)

(x-5)(x+2)

x(x-6)

Correct answer:

(x-6)(x-1)

Explanation:

We will discuss coefficients in the general equation:

ax^2 +bx+c

In this case, is positive and is negative, and a=1 , so we know our answer involves two negative numbers that are factors of and add to . The answer is:

(x-6)(x-1)

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Example Question #1 : How To Factor A Polynomial

Solve for : $x^{2}+16x=-28$

Possible Answers:

x=-4,-7

x=-2,-14

x=1,28

x=2,14

Correct answer:

x=-2,-14

Explanation:

This is a factoring problem so we need to get all of the variables on one side and set the equation equal to zero. To do this we subtract from both sides to get $x^{2}+16x+28=0$

Think of the equation in this format to help with the following explanation. $ax^{2}+bx+c=0$

We must then factor to find the solutions for . To do this we must make a factor tree of which is 28 in this case to find the possible solutions. The possible numbers are $1\cdot 28$ , $2\cdot 14$ , $4\cdot 7$ .

Since is positive we know that our factoring will produce two positive numbers.

We then use addition with the factoring tree to find the numbers that add together to equal . So 28+1=29 , 2+14=16 , and 4+7=11

Success! 14 plus 2 equals . We then plug our numbers into the factored form of (x+2)(x+14)=0

We know that anything multiplied by 0 is equal to 0 so we plug in the numbers for which make each equation equal to 0 so in this case x=-2,-14 .

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Example Question #1 : How To Factor A Polynomial

Solve by factoring:

$x^{2} + 2x - 15$

Possible Answers:

x = -3, 5

x = 15

x = 5, -5

x= 3, -3

x = 3, -5

Correct answer:

x = 3, -5

Explanation:

By factoring one gets

$y = (x-3)\left ( x + 5 \right )$

Now setting each of the two factors to 0 (using the zero property), one gets

x= 3 or x= -5

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Example Question #3 : How To Factor A Polynomial

Solve by using the quadratic formula:

$6x^{2}+x +15$

Possible Answers:

$-1 + \sqrt{359}i$

$-1 - \sqrt{359}i$

$\frac{1}{12} - \frac{\sqrt{359}}{12}$

$\frac{-1}{12}+ \frac{\sqrt{359}i}{12}$

$\frac{-1}{12} - \frac{\sqrt{359 }i}{12}$

$\frac{1}{12} + \frac{\sqrt{359}}{12}$

$\frac{-1}{12} + \frac{\sqrt{359}}{12}$

Correct answer:

$\frac{-1}{12}+ \frac{\sqrt{359}i}{12}$

$\frac{-1}{12} - \frac{\sqrt{359 }i}{12}$

Explanation:

For the quadeatic equation $a=6,b=1,and\ c=15$ . Applying these to the quadratic formula

$\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

we get $\frac{-1 \pm \sqrt{1-4(6)15)}}{2\cdot 6}$

resulting in

$\frac{-1 \pm \sqrt{-359}}{12} = \frac{-1 +\sqrt{359}i}{12}$

and $\frac{-1 - \sqrt{359}i}{12}$

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Example Question #4551 : Algebra 1

Find the axis of symmetry and the minimum/maximum value of the following quadratic equation in standard form:

$- x^{2} + 4x - 12$

Possible Answers:

x=2 , Maximum value of the function =-8

x=-2 , Minimum value of the function =-8

x=2 , Minimum value of the function =-8

x=2 , Minimum value of the function

x=-2 , Maximum value of the function =-8

Correct answer:

x=2 , Maximum value of the function =-8

Explanation:

If we convert the given quadratic equation from standard form to vertex form, we get:

$-\left ( x-2 \right )^{2} -8$

Hence the axis of symmetry is

x=2 and the minimum value at x=2 is ( this function is concave down)

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Example Question #5 : Solving By Factoring

Write the equation of a circle having (3, 4) as center and a radius of $\sqrt{5}$ .

Possible Answers:

$\left ( x-3 \right )^{2} + \left ( y-4 \right )^{2} = \sqrt{5}$

$\left ( x-3 \right )^{2} + \left ( y+4 \right )^{2} =\sqrt{5}$

$\left ( x-3 \right )^{3} + \left ( y +3 \right )^{2} = 5$

$\left ( x-3 \right )^{2} + \left ( y-4 \right )^{2} = 5$

$\left ( x+3) \right )^{2} + \left ( y-4 \right )^{2} = 5$

Correct answer:

$\left ( x-3 \right )^{2} + \left ( y-4 \right )^{2} = 5$

Explanation:

The center is located at (3,4) which means the standard equation of a circle which is:

$x^{2} +y^{2} = r^{2}$

becomes

$\left ( x-3 \right )^{2} + \left ( y-4 \right )^{2} = \sqrt{5}^{2}$

which equals to

$\left ( x-3 \right )^{2}+\left ( y-4 \right )^{2} = 5$

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Example Question #1 : Factoring Polynomials

Factor:

$8x^{3} - 1$

Possible Answers:

$\left ( 2x^{2} \right +1)\left ( 4x-1 \right )$

$\left ( 2x-1 \right )\left \left ( 4x^{2} \right + 2x + 1)$

$\left ( 2x + 1 \right )\left ( 4x^{2} \right -1 )$

$\left ( 2x-1 \right )\left ( 4x^{2} \right +1)$

$\left ( 2x-1 \right )^{3}$

Correct answer:

$\left ( 2x-1 \right )\left \left ( 4x^{2} \right + 2x + 1)$

Explanation:

This is a difference of cubes.

$8,x^{3},\ and\ -1$ are all cubes. So the formula for factoring this expression is:

$a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$

$8x^{3}-1=(2x-1)(4x^{2}+2x+1)$

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Example Question #6 : Factoring Polynomials

Equation of a circle in standard form:

$\left ( x-h \right )^{2} + \left ( y - k \right )^{2} = r^{2}$

with center $\left ( h,k \right )$ and radius equal to .

Find out the radius and center of circle from the given equation in expanded form:

$x^{2} +2x + y^{2} -4x = 6$

Possible Answers:

The center is $\left ( -1,2 \right )$ and the radius = $\sqrt{11}$

Center : $\left ( 1,-2 \right )$ and the radius = $\sqrt{11}$

Center: $\left ( 1,2 \right )$ and the radius = $\sqrt{11}$

Center: $\left ( 1,-2 \right )$ and the radius = $\sqrt{6}$

Center: $\left ( -1,-2 \right )$ and the radius = $\sqrt{11}$

Correct answer:

The center is $\left ( -1,2 \right )$ and the radius = $\sqrt{11}$

Explanation:

The given equation in standard form :

$\left ( x+1 \right )^{2} + \left ( y-2 \right )^{2} = 11$

Hence the center: $\left ( -1,2 \right )$ and the radius = $\sqrt{11}$ .

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Example Question #2 : How To Factor A Polynomial

Factor:

$9x^{2} - 4y^{2}$

Possible Answers:

$\left ( 3x-2y \right )\left ( 3x-2y \right )$

$\left ( 3x+2y \right )\left ( 3x+2y \right )$

$\left ( 3x + 2y \right )^{2}$

$\left ( 3x + 2y \right )\left ( 3x - 2y \right )$

$\left ( 3x - 2y \right )^{2}$

Correct answer:

$\left ( 3x + 2y \right )\left ( 3x - 2y \right )$

Explanation:

Difference of 2 squares is equal to the product of the sum and difference of two terms.

The two terms here are and and the product therefore equals to

$\left ( 3x-2y \right )\left ( 3x + 2y \right )$

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Example Question #3 : Factoring Polynomials

Simplify:

$\frac{x+1}{x^{2}-1}\cdot$ $\frac{x^{2}+2x-3}{2x}$

Possible Answers:

$\frac{x+1}{2x}$

$\frac{x+3}{x^{2}-1}$

$\frac{\left ( x+3 \right )\left ( x+1 \right )}{\left ( x-1 \right )2x}$

$\frac{x+3}{2x}$

$\frac{x+3}{x^{2}-1}$

Correct answer:

$\frac{x+3}{2x}$

Explanation:

By factoring both the numerator and the denominator we get the following:

$\frac{\left ( x+1 \right )\left ( x-1 \right )\left ( x+3 \right )}{\left ( x+1 \right )\left ( x-1 \right )2x}$

If we simplify we get:

$\frac{x+3}{2x}$

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Algebra 1 : Factoring Polynomials

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #1 : How To Factor A Polynomial

Example Question #1 : How To Factor A Polynomial

Example Question #1 : How To Factor A Polynomial

Example Question #3 : How To Factor A Polynomial

Example Question #4551 : Algebra 1

Example Question #5 : Solving By Factoring

Example Question #1 : Factoring Polynomials

Example Question #6 : Factoring Polynomials

Example Question #2 : How To Factor A Polynomial

Example Question #3 : Factoring Polynomials

All Algebra 1 Resources

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