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Common Core: High School - Algebra : Arithmetic with Polynomials & Rational Expressions

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All Common Core: High School - Algebra Resources

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Example Questions

Example Question #233 : High School: Algebra

Find the zeros of $- 5 x^{2} + 36 x + 76$ $\displaystyle - 5 x^{2} + 36 x + 76$

Possible Answers:

$\displaystyle -36 , 36$

There are no real roots

$\frac{18}{5} + \frac{8 \sqrt{11}}{5} , - \frac{8 \sqrt{11}}{5} + \frac{18}{5}$ $\displaystyle \frac{18}{5} + \frac{8 \sqrt{11}}{5} , - \frac{8 \sqrt{11}}{5} + \frac{18}{5}$

$\displaystyle 0$

$16 \sqrt{11} , - 16 \sqrt{11}$ $\displaystyle 16 \sqrt{11} , - 16 \sqrt{11}$

Correct answer:

$\frac{18}{5} + \frac{8 \sqrt{11}}{5} , - \frac{8 \sqrt{11}}{5} + \frac{18}{5}$ $\displaystyle \frac{18}{5} + \frac{8 \sqrt{11}}{5} , - \frac{8 \sqrt{11}}{5} + \frac{18}{5}$

Explanation:

In order to find the zeros, we can use the quadratic formula.

Recall the quadratic formula.

$x =\frac{ - b \pm \sqrt{- 4 a c + b^{2}} }{ 2 a }$ $\displaystyle x =\frac{ - b \pm \sqrt{- 4 a c + b^{2}} }{ 2 a }$

Where $\uptext{a}$ $\displaystyle \uptext{a}$, $\uptext{b}$ $\displaystyle \uptext{b}$, and $\uptext{c}$ $\displaystyle \uptext{c}$, correspond to the coefficients in the equation

$a x^{2} + b x + c = 0$ $\displaystyle a x^{2} + b x + c = 0$

In this case a = -5 $\displaystyle a = -5$, b = 36 $\displaystyle b = 36$ and c = 76 $\displaystyle c = 76$

We plug in these values into the quadratic formula, and evaluate them.

$x =\frac{ -36 \pm \sqrt{ 36 ^2 - 4 * -5 * 76 }}{ 2 * -5 }$ $\displaystyle x =\frac{ -36 \pm \sqrt{ 36 ^2 - 4 * -5 * 76 }}{ 2 * -5 }$

$x =\frac{ -36 \pm \sqrt{ 1296 - -1520 }}{ -10 }$ $\displaystyle x =\frac{ -36 \pm \sqrt{ 1296 - -1520 }}{ -10 }$

$x =\frac{ -36 \pm \sqrt{ 2816 }}{ -10 }$ $\displaystyle x =\frac{ -36 \pm \sqrt{ 2816 }}{ -10 }$

$x =\frac{ -36 \pm 16 \sqrt{11} }{ -10 }$ $\displaystyle x =\frac{ -36 \pm 16 \sqrt{11} }{ -10 }$

Now we split this up into two equations.

$x =\frac{ -36 + 16 \sqrt{11} }{ -10 }$ $\displaystyle x =\frac{ -36 + 16 \sqrt{11} }{ -10 }$

$x =\frac{ -36 - 16 \sqrt{11} }{ -10 }$ $\displaystyle x =\frac{ -36 - 16 \sqrt{11} }{ -10 }$

So our zeros are at

$x = \frac{18}{5} + \frac{8 \sqrt{11}}{5}$ $\displaystyle x = \frac{18}{5} + \frac{8 \sqrt{11}}{5}$

$x = - \frac{8 \sqrt{11}}{5} + \frac{18}{5}$ $\displaystyle x = - \frac{8 \sqrt{11}}{5} + \frac{18}{5}$

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Example Question #234 : High School: Algebra

Find the zeros of $4 x^{2} + 29 x + 19$ $\displaystyle 4 x^{2} + 29 x + 19$

Possible Answers:

$\sqrt{537} , - \sqrt{537}$ $\displaystyle \sqrt{537} , - \sqrt{537}$

There are no real roots

$- \frac{29}{8} - \frac{\sqrt{537}}{8} , - \frac{29}{8} + \frac{\sqrt{537}}{8}$ $\displaystyle - \frac{29}{8} - \frac{\sqrt{537}}{8} , - \frac{29}{8} + \frac{\sqrt{537}}{8}$

$\displaystyle 0$

$\displaystyle -29 , 29$

Correct answer:

$- \frac{29}{8} - \frac{\sqrt{537}}{8} , - \frac{29}{8} + \frac{\sqrt{537}}{8}$ $\displaystyle - \frac{29}{8} - \frac{\sqrt{537}}{8} , - \frac{29}{8} + \frac{\sqrt{537}}{8}$

Explanation:

In order to find the zeros, we can use the quadratic formula.

Recall the quadratic formula.

$a x^{2} + b x + c = 0$ $\displaystyle a x^{2} + b x + c = 0$

In this case a = 4 $\displaystyle a = 4$, b = 29 $\displaystyle b = 29$, and c = 19 $\displaystyle c = 19$

We plug in these values into the quadratic formula, and evaluate them.

$x =\frac{ -29 \pm \sqrt{ 29 ^2 - 4 * 4 * 19 }}{ 2 * 4 }$ $\displaystyle x =\frac{ -29 \pm \sqrt{ 29 ^2 - 4 * 4 * 19 }}{ 2 * 4 }$

$x =\frac{ -29 \pm \sqrt{ 841 - 304 }}{ 8 }$ $\displaystyle x =\frac{ -29 \pm \sqrt{ 841 - 304 }}{ 8 }$

$x =\frac{ -29 \pm \sqrt{ 537 }}{ 8 }$ $\displaystyle x =\frac{ -29 \pm \sqrt{ 537 }}{ 8 }$

$x =\frac{ -29 \pm \sqrt{537} }{ 8 }$ $\displaystyle x =\frac{ -29 \pm \sqrt{537} }{ 8 }$

Now we split this up into two equations.

$x =\frac{ -29 + \sqrt{537} }{ 8 }$ $\displaystyle x =\frac{ -29 + \sqrt{537} }{ 8 }$

$x =\frac{ -29 - \sqrt{537} }{ 8 }$ $\displaystyle x =\frac{ -29 - \sqrt{537} }{ 8 }$

So our zeros are at

$x = - \frac{29}{8} - \frac{\sqrt{537}}{8}$ $\displaystyle x = - \frac{29}{8} - \frac{\sqrt{537}}{8}$

$x = - \frac{29}{8} + \frac{\sqrt{537}}{8}$ $\displaystyle x = - \frac{29}{8} + \frac{\sqrt{537}}{8}$

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Example Question #235 : High School: Algebra

Find the zeros of $x^{2} + 34 x + 24$ $\displaystyle x^{2} + 34 x + 24$

Possible Answers:

$-17 - \sqrt{265} , -17 + \sqrt{265}$ $\displaystyle -17 - \sqrt{265} , -17 + \sqrt{265}$

$\displaystyle -34 , 34$

$2 \sqrt{265} , - 2 \sqrt{265}$ $\displaystyle 2 \sqrt{265} , - 2 \sqrt{265}$

$\displaystyle 0$

There are no real roots

Correct answer:

$-17 - \sqrt{265} , -17 + \sqrt{265}$ $\displaystyle -17 - \sqrt{265} , -17 + \sqrt{265}$

Explanation:

In order to find the zeros, we can use the quadratic formula.

Recall the quadratic formula.

$x =\frac{ - b \pm \sqrt{- 4 a c + b^{2}} }{ 2 a }$ $\displaystyle x =\frac{ - b \pm \sqrt{- 4 a c + b^{2}} }{ 2 a }$

Where $\uptext{a}$ $\displaystyle \uptext{a}$, $\uptext{b}$ $\displaystyle \uptext{b}$, and $\uptext{c}$ $\displaystyle \uptext{c}$, correspond to the coefficients in the equation

$a x^{2} + b x + c = 0$ $\displaystyle a x^{2} + b x + c = 0$

In this case a = 1 $\displaystyle a = 1$, b = 34 $\displaystyle b = 34$ and c = 24 $\displaystyle c = 24$

We plug in these values into the quadratic formula, and evaluate them.

$x =\frac{ -34 \pm \sqrt{ 34 ^2 - 4 * 1 * 24 }}{ 2 * 1 }$ $\displaystyle x =\frac{ -34 \pm \sqrt{ 34 ^2 - 4 * 1 * 24 }}{ 2 * 1 }$

$x =\frac{ -34 \pm \sqrt{ 1156 - 96 }}{ 2 }$ $\displaystyle x =\frac{ -34 \pm \sqrt{ 1156 - 96 }}{ 2 }$

$x =\frac{ -34 \pm \sqrt{ 1060 }}{ 2 }$ $\displaystyle x =\frac{ -34 \pm \sqrt{ 1060 }}{ 2 }$

$x =\frac{ -34 \pm 2 \sqrt{265} }{ 2 }$ $\displaystyle x =\frac{ -34 \pm 2 \sqrt{265} }{ 2 }$

Now we split this up into two equations.

$x =\frac{ -34 + 2 \sqrt{265} }{ 2 }$ $\displaystyle x =\frac{ -34 + 2 \sqrt{265} }{ 2 }$

$x =\frac{ -34 - 2 \sqrt{265} }{ 2 }$ $\displaystyle x =\frac{ -34 - 2 \sqrt{265} }{ 2 }$

So our zeros are at

$\\x = -17 - \sqrt{265} \\x = -17 + \sqrt{265}$ $\displaystyle \\x = -17 - \sqrt{265} \\x = -17 + \sqrt{265}$

Report an Error

Example Question #236 : High School: Algebra

Find the zeros of $- 5 x^{2} + 30 x - 16$ $\displaystyle - 5 x^{2} + 30 x - 16$

Possible Answers:

$- \frac{\sqrt{145}}{5} + 3 , \frac{\sqrt{145}}{5} + 3$ $\displaystyle - \frac{\sqrt{145}}{5} + 3 , \frac{\sqrt{145}}{5} + 3$

$\displaystyle -30 , 30$

$\displaystyle 0$

There are no real roots

$2 \sqrt{145} , - 2 \sqrt{145}$ $\displaystyle 2 \sqrt{145} , - 2 \sqrt{145}$

Correct answer:

$- \frac{\sqrt{145}}{5} + 3 , \frac{\sqrt{145}}{5} + 3$ $\displaystyle - \frac{\sqrt{145}}{5} + 3 , \frac{\sqrt{145}}{5} + 3$

Explanation:

In order to find the zeros, we can use the quadratic formula.

Recall the quadratic formula.

$a x^{2} + b x + c = 0$ $\displaystyle a x^{2} + b x + c = 0$

In this case a = -5 $\displaystyle a = -5$, b = 30 $\displaystyle b = 30$ and c = -16 $\displaystyle c = -16$

We plug in these values into the quadratic formula, and evaluate them.

$x =\frac{ -30 \pm \sqrt{ 30 ^2 - 4 * -5 * -16 }}{ 2 * -5 }$ $\displaystyle x =\frac{ -30 \pm \sqrt{ 30 ^2 - 4 * -5 * -16 }}{ 2 * -5 }$

$x =\frac{ -30 \pm \sqrt{ 900 - 320 }}{ -10 }$ $\displaystyle x =\frac{ -30 \pm \sqrt{ 900 - 320 }}{ -10 }$

$x =\frac{ -30 \pm \sqrt{ 580 }}{ -10 }$ $\displaystyle x =\frac{ -30 \pm \sqrt{ 580 }}{ -10 }$

$x =\frac{ -30 \pm 2 \sqrt{145} }{ -10 }$ $\displaystyle x =\frac{ -30 \pm 2 \sqrt{145} }{ -10 }$

Now we split this up into two equations.

$x =\frac{ -30 + 2 \sqrt{145} }{ -10 }$ $\displaystyle x =\frac{ -30 + 2 \sqrt{145} }{ -10 }$

$x =\frac{ -30 - 2 \sqrt{145} }{ -10 }$ $\displaystyle x =\frac{ -30 - 2 \sqrt{145} }{ -10 }$

So our zeros are at

$x = - \frac{\sqrt{145}}{5} + 3$ $\displaystyle x = - \frac{\sqrt{145}}{5} + 3$

$x = \frac{\sqrt{145}}{5} + 3$ $\displaystyle x = \frac{\sqrt{145}}{5} + 3$

Report an Error

Example Question #237 : High School: Algebra

Find the zeros of $- 3 x^{2} + 24 x + 87$ $\displaystyle - 3 x^{2} + 24 x + 87$

Possible Answers:

$4 + 3 \sqrt{5} , - 3 \sqrt{5} + 4$ $\displaystyle 4 + 3 \sqrt{5} , - 3 \sqrt{5} + 4$

$\displaystyle -24 , 24$

$\displaystyle 0$

$18 \sqrt{5} , - 18 \sqrt{5}$ $\displaystyle 18 \sqrt{5} , - 18 \sqrt{5}$

There are no real roots

Correct answer:

$4 + 3 \sqrt{5} , - 3 \sqrt{5} + 4$ $\displaystyle 4 + 3 \sqrt{5} , - 3 \sqrt{5} + 4$

Explanation:

$a x^{2} + b x + c = 0$ $\displaystyle a x^{2} + b x + c = 0$

In this case a = -3 $\displaystyle a = -3$, b = 24 $\displaystyle b = 24$, and c = 87 $\displaystyle c = 87$

We plug in these values into the quadratic formula, and evaluate them.

$x =\frac{ -24 \pm \sqrt{ 24 ^2 - 4 * -3 * 87 }}{ 2 * -3 }$ $\displaystyle x =\frac{ -24 \pm \sqrt{ 24 ^2 - 4 * -3 * 87 }}{ 2 * -3 }$

$x =\frac{ -24 \pm \sqrt{ 576 - -1044 }}{ -6 }$ $\displaystyle x =\frac{ -24 \pm \sqrt{ 576 - -1044 }}{ -6 }$

$x =\frac{ -24 \pm \sqrt{ 1620 }}{ -6 }$ $\displaystyle x =\frac{ -24 \pm \sqrt{ 1620 }}{ -6 }$

$x =\frac{ -24 \pm 18 \sqrt{5} }{ -6 }$ $\displaystyle x =\frac{ -24 \pm 18 \sqrt{5} }{ -6 }$

Now we split this up into two equations.

$x =\frac{ -24 + 18 \sqrt{5} }{ -6 }$ $\displaystyle x =\frac{ -24 + 18 \sqrt{5} }{ -6 }$

$x =\frac{ -24 - 18 \sqrt{5} }{ -6 }$ $\displaystyle x =\frac{ -24 - 18 \sqrt{5} }{ -6 }$

So our zeros are at

$x = 4 + 3 \sqrt{5}$ $\displaystyle x = 4 + 3 \sqrt{5}$

$x = - 3 \sqrt{5} + 4$ $\displaystyle x = - 3 \sqrt{5} + 4$

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Example Question #238 : High School: Algebra

Find the zeros of $- x^{2} + 18 x - 53$ $\displaystyle - x^{2} + 18 x - 53$

Possible Answers:

$\displaystyle -18 , 18$

$- 2 \sqrt{7} + 9 , 2 \sqrt{7} + 9$ $\displaystyle - 2 \sqrt{7} + 9 , 2 \sqrt{7} + 9$

There are no real roots

$\displaystyle 0$

$4 \sqrt{7} , - 4 \sqrt{7}$ $\displaystyle 4 \sqrt{7} , - 4 \sqrt{7}$

Correct answer:

$- 2 \sqrt{7} + 9 , 2 \sqrt{7} + 9$ $\displaystyle - 2 \sqrt{7} + 9 , 2 \sqrt{7} + 9$

Explanation:

$a x^{2} + b x + c = 0$ $\displaystyle a x^{2} + b x + c = 0$

In this case a = -1 $\displaystyle a = -1$, b = 18 $\displaystyle b = 18$, and c = -53 $\displaystyle c = -53$

We plug in these values into the quadratic formula, and evaluate them.

$x =\frac{ -18 \pm \sqrt{ 18 ^2 - 4 * -1 * -53 }}{ 2 * -1 }$ $\displaystyle x =\frac{ -18 \pm \sqrt{ 18 ^2 - 4 * -1 * -53 }}{ 2 * -1 }$

$x =\frac{ -18 \pm \sqrt{ 324 - 212 }}{ -2 }$ $\displaystyle x =\frac{ -18 \pm \sqrt{ 324 - 212 }}{ -2 }$

$x =\frac{ -18 \pm \sqrt{ 112 }}{ -2 }$ $\displaystyle x =\frac{ -18 \pm \sqrt{ 112 }}{ -2 }$

$x =\frac{ -18 \pm 4 \sqrt{7} }{ -2 }$ $\displaystyle x =\frac{ -18 \pm 4 \sqrt{7} }{ -2 }$

Now we split this up into two equations.

$x =\frac{ -18 + 4 \sqrt{7} }{ -2 }$ $\displaystyle x =\frac{ -18 + 4 \sqrt{7} }{ -2 }$

$x =\frac{ -18 - 4 \sqrt{7} }{ -2 }$ $\displaystyle x =\frac{ -18 - 4 \sqrt{7} }{ -2 }$

So our zeros are at

$x = - 2 \sqrt{7} + 9$ $\displaystyle x = - 2 \sqrt{7} + 9$

$x = 2 \sqrt{7} + 9$ $\displaystyle x = 2 \sqrt{7} + 9$

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Example Question #239 : High School: Algebra

Find the zeros of $- 7 x^{2} + 34 x + 16$ $\displaystyle - 7 x^{2} + 34 x + 16$

Possible Answers:

There are no real roots

$2 \sqrt{401} , - 2 \sqrt{401}$ $\displaystyle 2 \sqrt{401} , - 2 \sqrt{401}$

$\frac{17}{7} + \frac{\sqrt{401}}{7} , - \frac{\sqrt{401}}{7} + \frac{17}{7}$ $\displaystyle \frac{17}{7} + \frac{\sqrt{401}}{7} , - \frac{\sqrt{401}}{7} + \frac{17}{7}$

$\displaystyle 0$

$\displaystyle -34 , 34$

Correct answer:

$\frac{17}{7} + \frac{\sqrt{401}}{7} , - \frac{\sqrt{401}}{7} + \frac{17}{7}$ $\displaystyle \frac{17}{7} + \frac{\sqrt{401}}{7} , - \frac{\sqrt{401}}{7} + \frac{17}{7}$

Explanation:

In order to find the zeros, we can use the quadratic formula.

Recall the quadratic formula.

$a x^{2} + b x + c = 0$ $\displaystyle a x^{2} + b x + c = 0$

In this case a = -7 $\displaystyle a = -7$, b = 34 $\displaystyle b = 34$, and c = 16 $\displaystyle c = 16$

We plug in these values into the quadratic formula, and evaluate them.

$x =\frac{ -34 \pm \sqrt{ 34 ^2 - 4 * -7 * 16 }}{ 2 * -7 }$ $\displaystyle x =\frac{ -34 \pm \sqrt{ 34 ^2 - 4 * -7 * 16 }}{ 2 * -7 }$

$x =\frac{ -34 \pm \sqrt{ 1156 - -448 }}{ -14 }$ $\displaystyle x =\frac{ -34 \pm \sqrt{ 1156 - -448 }}{ -14 }$

$x =\frac{ -34 \pm \sqrt{ 1604 }}{ -14 }$ $\displaystyle x =\frac{ -34 \pm \sqrt{ 1604 }}{ -14 }$

$x =\frac{ -34 \pm 2 \sqrt{401} }{ -14 }$ $\displaystyle x =\frac{ -34 \pm 2 \sqrt{401} }{ -14 }$

Now we split this up into two equations.

$x =\frac{ -34 + 2 \sqrt{401} }{ -14 }$ $\displaystyle x =\frac{ -34 + 2 \sqrt{401} }{ -14 }$

$x =\frac{ -34 - 2 \sqrt{401} }{ -14 }$ $\displaystyle x =\frac{ -34 - 2 \sqrt{401} }{ -14 }$

So our zeros are at

$x = \frac{17}{7} + \frac{\sqrt{401}}{7}$ $\displaystyle x = \frac{17}{7} + \frac{\sqrt{401}}{7}$

$x = - \frac{\sqrt{401}}{7} + \frac{17}{7}$ $\displaystyle x = - \frac{\sqrt{401}}{7} + \frac{17}{7}$

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Example Question #240 : High School: Algebra

Find the zeros of $5 x^{2} + 17 x - 21$ $\displaystyle 5 x^{2} + 17 x - 21$

Possible Answers:

$\sqrt{709} , - \sqrt{709}$ $\displaystyle \sqrt{709} , - \sqrt{709}$

There are no real roots

$- \frac{17}{10} + \frac{\sqrt{709}}{10} , - \frac{\sqrt{709}}{10} - \frac{17}{10}$ $\displaystyle - \frac{17}{10} + \frac{\sqrt{709}}{10} , - \frac{\sqrt{709}}{10} - \frac{17}{10}$

$\displaystyle 0$

$\displaystyle -17 , 17$

Correct answer:

$- \frac{17}{10} + \frac{\sqrt{709}}{10} , - \frac{\sqrt{709}}{10} - \frac{17}{10}$ $\displaystyle - \frac{17}{10} + \frac{\sqrt{709}}{10} , - \frac{\sqrt{709}}{10} - \frac{17}{10}$

Explanation:

$a x^{2} + b x + c = 0$ $\displaystyle a x^{2} + b x + c = 0$

In this case a = 5 $\displaystyle a = 5$, b = 17 $\displaystyle b = 17$, and c = -21 $\displaystyle c = -21$.

We plug in these values into the quadratic formula, and evaluate them.

$x =\frac{ -17 \pm \sqrt{ 17 ^2 - 4 * 5 * -21 }}{ 2 * 5 }$ $\displaystyle x =\frac{ -17 \pm \sqrt{ 17 ^2 - 4 * 5 * -21 }}{ 2 * 5 }$

$x =\frac{ -17 \pm \sqrt{ 289 - -420 }}{ 10 }$ $\displaystyle x =\frac{ -17 \pm \sqrt{ 289 - -420 }}{ 10 }$

$x =\frac{ -17 \pm \sqrt{ 709 }}{ 10 }$ $\displaystyle x =\frac{ -17 \pm \sqrt{ 709 }}{ 10 }$

Now we split this up into two equations.

$x =\frac{ -17 + \sqrt{709} }{ 10 }$ $\displaystyle x =\frac{ -17 + \sqrt{709} }{ 10 }$

$x =\frac{ -17 - \sqrt{709} }{ 10 }$ $\displaystyle x =\frac{ -17 - \sqrt{709} }{ 10 }$

So our zeros are at

$x = - \frac{17}{10} + \frac{\sqrt{709}}{10}$ $\displaystyle x = - \frac{17}{10} + \frac{\sqrt{709}}{10}$

$x = - \frac{\sqrt{709}}{10} - \frac{17}{10}$ $\displaystyle x = - \frac{\sqrt{709}}{10} - \frac{17}{10}$

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Example Question #21 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

What are the $\displaystyle x$-intercept(s) of the function?

$\displaystyle y=x^2+10x+25$

Possible Answers:

$\text{x-intercepts: }x=-5 \text{ and } x=5$ $\displaystyle \text{x-intercepts: }x=-5 \text{ and } x=5$

$\text{x-intercepts: }x=-5 \text{ and } x=0$ $\displaystyle \text{x-intercepts: }x=-5 \text{ and } x=0$

$\text{x-intercepts: }x=-5 \text{ and } x=-1$ $\displaystyle \text{x-intercepts: }x=-5 \text{ and } x=-1$

$\text{x-intercept: }x=-5$ $\displaystyle \text{x-intercept: }x=-5$

$\text{x-intercept: }x=5$ $\displaystyle \text{x-intercept: }x=5$

Correct answer:

$\text{x-intercept: }x=-5$ $\displaystyle \text{x-intercept: }x=-5$

Explanation:

To find the $\displaystyle x$-intercept of a function, first recall that the $\displaystyle x$-intercept represents the points where the graph of the function crosses the $\displaystyle x$-axis. In other words where the function has a $\displaystyle y$ value equal to zero.

One technique that can be used is factorization. In general form,

$y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)$ $\displaystyle y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)$

where,

c_1 $\displaystyle c_1$ and c_2 $\displaystyle c_2$ are factors of $\displaystyle c$ and when added together results in $\displaystyle b$.

For the given function,

$\displaystyle y=x^2+10x+25$

the coefficients are,

$\\b=10 \\c=25$ $\displaystyle \\b=10 \\c=25$

therefore the factors of $\displaystyle c$ that have a sum of $\displaystyle b$ are,

$\displaystyle y=(x+5)(x+5)$

Now find the $\displaystyle x$-intercepts of the function by setting each binomial equal to zero and solving for $\displaystyle x$.

$\\x+5=0\rightarrow x=-5$ $\displaystyle \\x+5=0\rightarrow x=-5$

To verify, graph the function.

Screen shot 2016 03 09 at 10.24.54 am

The graph crosses the $\displaystyle x$-axis at -5, thus verifying the result found by factorization.

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Example Question #1 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Use FOIL for the following expression.

$\left(a + 10 b\right)^{2}$ $\displaystyle \left(a + 10 b\right)^{2}$

Possible Answers:

$a^{2}$ $\displaystyle a^{2}$

$b^{2}$ $\displaystyle b^{2}$

$10 a b + 100 b^{2}$ $\displaystyle 10 a b + 100 b^{2}$

$a^{2} + 20 a b + 100 b^{2}$ $\displaystyle a^{2} + 20 a b + 100 b^{2}$

$a^{2} + 20 a b$ $\displaystyle a^{2} + 20 a b$

Correct answer:

$a^{2} + 20 a b + 100 b^{2}$ $\displaystyle a^{2} + 20 a b + 100 b^{2}$

Explanation:

The first step is to rewrite the problem as follows.

$\left(a + 10 b\right)^{2} = \left( a + 10*b \right) \cdot \left( a + 10*b \right)$ $\displaystyle \left(a + 10 b\right)^{2} = \left( a + 10*b \right) \cdot \left( a + 10*b \right)$

Now we multiply the first parts of the first and second expression together.

$a \cdot a = a^{2}$ $\displaystyle a \cdot a = a^{2}$

Now we multiply the first term of the first expression with the second term of the second expression.

$a \cdot 10 b = 10 a b$ $\displaystyle a \cdot 10 b = 10 a b$

Now we multiply the second term of the first expression with the first term of the second expression.

$a \cdot 10 b = 10 a b$ $\displaystyle a \cdot 10 b = 10 a b$

Now we multiply the last terms of each expression together.

$10 b \cdot 10 b = 100 b^{2}$ $\displaystyle 10 b \cdot 10 b = 100 b^{2}$

Now we add all these results together, and we get.

$a^{2} + 20 a b + 100 b^{2}$ $\displaystyle a^{2} + 20 a b + 100 b^{2}$

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All Common Core: High School - Algebra Resources

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Common Core: High School - Algebra : Arithmetic with Polynomials & Rational Expressions

Study concepts, example questions & explanations for Common Core: High School - Algebra

All Common Core: High School - Algebra Resources

Example Questions

Example Question #233 : High School: Algebra

Example Question #234 : High School: Algebra

Example Question #235 : High School: Algebra

Example Question #236 : High School: Algebra

Example Question #237 : High School: Algebra

Example Question #238 : High School: Algebra

Example Question #239 : High School: Algebra

Example Question #240 : High School: Algebra

Example Question #21 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

Example Question #1 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

All Common Core: High School - Algebra Resources

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