Solving Trinomials - Examples, Exercises and Solutions

Proceed to solve the given equation:

$x^2+10x=-25$

First, let's arrange the equation by moving terms:

$x^2+10x=-25 \\ x^2+10x+25=0 \\$Note that the expression on the left side can be factored using the perfect square trinomial formula for a binomial squared:

$(\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$

As shown below:

$25=5^2$

Therefore, we'll represent the rightmost term as a squared term:

$x^2+10x+25=0 \\ \downarrow\\ \textcolor{red}{x}^2+10x+\textcolor{blue}{5}^2=0$

Now let's examine again the perfect square trinomial formula mentioned earlier:

$(\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

And the expression on the left side in the equation that we obtained in the last step:

$\textcolor{red}{x}^2+\underline{10x}+\textcolor{blue}{5}^2=0$

Notice that the terms $\textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{5}^2$indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor this expression (on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined with a line):

$(\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

In other words - we will query whether we can represent the expression on the left side as:

$\textcolor{red}{x}^2+\underline{10x}+\textcolor{blue}{5}^2=0 \\ \updownarrow\text{?}\\ \textcolor{red}{x}^2+\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}}+\textcolor{blue}{5}^2=0$

And indeed it is true that:

$2\cdot x\cdot5=10x$

Therefore we can represent the expression on the left side of the equation as a perfect square binomial:

$\textcolor{red}{x}^2+2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{5})^2=0$

From here we can take the square root of both sides of the equation (and don't forget there are two possibilities - positive and negative when taking the square root of an even power), then we'll easily solve by isolating the variable:

$(x+5)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x+5=\pm0\\ x+5=0\\ \boxed{x=-5}$

Let's summarize the solution of the equation:

$x^2+10x=-25 \\ x^2+10x+25=0 \\ \downarrow\\ \textcolor{red}{x}^2+2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{5})^2=0 \\ \downarrow\\ x+5=0\\ \downarrow\\ \boxed{x=-5}$

Therefore the correct answer is answer C.

Let's solve the given equation:

$60-16y+y^2=-4$First, let's arrange the equation by moving terms:

$60-16y+y^2=-4 \\ 60-16y+y^2+4=0 \\ y^2-16y+64=0$Now, let's note that we can break down the expression on the left side using the short quadratic factoring formula:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$This is done using the fact that:

$64=8^2$So let's present the outer term on the right as a square:

$y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-16y+\textcolor{blue}{8}^2=0$Now let's examine again the short factoring formula we mentioned earlier:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$And the expression on the left side of the equation we got in the last step:

$\textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2=0$Let's note that the terms $\textcolor{red}{y}^2,\hspace{6pt}\textcolor{blue}{8}^2$indeed match the form of the first and third terms in the short multiplication formula (which are highlighted in red and blue),

But in order for us to break down the relevant expression (which is on the left side of the equation) using the short formula we mentioned, the match to the short formula must also apply to the remaining term, meaning the middle term in the expression (underlined):

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$In other words - we'll ask if it's possible to present the expression on the left side of the equation as:

$\textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2 =0 \\ \updownarrow\text{?}\\ \textcolor{red}{y}^2-\underline{2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}}+\textcolor{blue}{8}^2 =0$And indeed it holds that:

$2\cdot y\cdot8=16y$So we can present the expression on the left side of the given equation as a difference of two squares:

$\textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0$From here we can take out square roots for the two sides of the equation (remember that there are two possibilities - positive and negative when taking out square roots), we'll solve it easily by isolating the variable on one side:

$(y-8)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ y-8=\pm0\\ y-8=0\\ \boxed{y=8}$

Let's summarize then the solution of the equation:

$60-16y+y^2=-4 \\ y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0 \\ \downarrow\\ y-8=0\\ \downarrow\\ \boxed{y=8}$

So the correct answer is answer a.

Solve the given equation:

$x^2-10x=-16$

First, let's arrange the equation by moving terms:

$x^2-10x=-16 \\ x^2-10x+16 =0$

Note that the coefficient of the squared term is 1, therefore, we can (try to) factor the expression on the left side by using quick trinomial factoring:

Let's look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=16\\ m+n=-10\\$From the first requirement, namely - the multiplication, we observe that the product of the numbers we're looking for must yield a positive result, therefore we can conclude that both numbers must have the same sign, according to multiplication rules. Remember that the possible factors of 16 are the number pairs 4 and 4, 2 and 8, or 16 and 1. Meeting the second requirement, along with the fact that the signs of the numbers we're looking for are identical leads us to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-8\\ n=-2 \end{cases}$

Therefore we'll factor the expression on the left side of the equation to:

$x^2-10x+16 =0 \\ \downarrow\\ (x-8)(x-2)=0$

From remember that the product of expressions will yield 0 only if at least one of the multiplied expressions equals zero,

Therefore we'll obtain two simple equations and solve them by isolating the unknown in each:

$x-8=0\\ \boxed{x=8}$

or:

$x-2=0\\ \boxed{x=2}$

Let's summarize the solution of the equation:

$x^2-10x=-16 \\ x^2-10x+16 =0 \\ \downarrow\\ (x-8)(x-2)=0 \\ \downarrow\\ x-8=0\rightarrow\boxed{x=8}\\ x-2=0\rightarrow\boxed{x=2}\\ \downarrow\\ \boxed{x=8,2}$

Therefore the correct answer is answer B.

Proceed to solve the given equation:

$x^2+144=24x$

Arrange the equation by moving terms:

$x^2+144=24x \\ x^2-24x+144=0$

Note that we are able to factor the expression on the left side by using the perfect square trinomial formula:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$

As demonstrated below:

$144=12^2$

Therefore, we'll represent the rightmost term as a squared term:

$x^2-24x+144=0 \\ \downarrow\\ \textcolor{red}{x}^2-24x+\textcolor{blue}{12}^2=0$

Now let's examine once again the perfect square trinomial formula mentioned earlier:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

And the expression on the left side in the equation that we obtained in the last step:

$\textcolor{red}{x}^2-\underline{24x}+\textcolor{blue}{12}^2=0$

Note that the terms $\textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{12}^2$indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor this expression (which is on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined):

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

In other words - we'll query whether we can represent the expression on the left side of the equation as:

$\textcolor{red}{x}^2-\underline{24x}+\textcolor{blue}{12}^2=0\\ \updownarrow\text{?}\\ \textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{12}}+\textcolor{blue}{12}^2=0$

And indeed it is true that:

$2\cdot x\cdot12=24x$

Therefore we can represent the expression on the left side of the equation as a perfect square trinomial:

$\textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{12}}+\textcolor{blue}{12}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{12})^2=0$

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable:

$(x-12)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x-12=\pm0\\ x-12=0\\ \boxed{x=12}$

Let's summarize the solution of the equation:

$x^2+144=24x \\ x^2-24x+144=0 \\ \downarrow\\ \textcolor{red}{x}^2-2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{12}+\textcolor{blue}{12}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{12})^2=0 \\ \downarrow\\ x-12=0\\ \downarrow\\ \boxed{x=12}$

Therefore the correct answer is answer C.

Proceed to solve the given equation:

$x^2=6x-9$

First, let's arrange the equation by moving terms:

$x^2=6x-9 \\ x^2-6x+9=0$

Note that we can factor the expression on the left side by using the perfect square trinomial formula for a binomial squared:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$

As shown below:

$9=3^2$Therefore, we'll represent the rightmost term as a squared term:

$x^2-6x+9=0 \\ \downarrow\\ \textcolor{red}{x}^2-6x+\textcolor{blue}{3}^2=0$

Now let's examine again the perfect square trinomial formula mentioned earlier:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

And the expression on the left side in the equation that we obtained in the last step:

$\textcolor{red}{x}^2-\underline{6x}+\textcolor{blue}{3}^2=0$

Note that the terms $\textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{3}^2$indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor the expression in question (which is on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined with a single line):

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

In other words - we will query whether we can represent the expression on the left side of the equation as:

$\textcolor{red}{x}^2-\underline{6x}+\textcolor{blue}{3}^2=0\\ \updownarrow\text{?}\\ \textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{3}}+\textcolor{blue}{3}^2=0$

And indeed it is true that:

$2\cdot x\cdot3=6x$

Therefore we can represent the expression on the left side of the equation as a perfect square binomial:

$\textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{3}}+\textcolor{blue}{3}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{3})^2=0$

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable:

$(x-3)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x-3=\pm0\\ x-3=0\\ \boxed{x=3}$

Let's summarize the solution of the equation:

$x^2=6x-9 \\ x^2-6x+9=0 \\ \downarrow\\ \textcolor{red}{x}^2-2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{3}+\textcolor{blue}{3}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{3})^2=0 \\ \downarrow\\ x-3=0\\ \downarrow\\ \boxed{x=3}$

Therefore the correct answer is answer C.

To solve the problem $x^4 - x^3 = 2x^2$, let's proceed as follows:

• Step 1: Set the equation to zero.
  $x^4 - x^3 - 2x^2 = 0$
• Step 2: Factor out the greatest common factor.
  The common factor among all terms is $x^2$.
  Factoring out $x^2$ gives:
  $x^2(x^2 - x - 2) = 0$
• Step 3: Solve the factors.
  This equation breaks into two factors that can be solved separately:
  • $x^2 = 0$
  • $x^2 - x - 2 = 0$
• Step 4: Solve $x^2 = 0$.
  Since $x^2 = 0$, we get:
  $x = 0$
• Step 5: Solve $x^2 - x - 2 = 0$.
  This can be factored further. We look for two numbers that multiply to $-2$ and add up to $-1$.
  These numbers are $-2$ and $1$, so we factor as:
  $(x - 2)(x + 1) = 0$
• Step 6: Solve the quadratic factors.
  Set each factor equal to zero:
  • $x - 2 = 0 \Rightarrow x = 2$
  • $x + 1 = 0 \Rightarrow x = -1$

The solutions to the equation $x^4 - x^3 = 2x^2$ are $x = -1, 0, 2$.

Therefore, the correct answer is:

$x = -1, 2, 0$

To solve the problem $x^3 = x^2 + 2x$, follow these steps:

• Step 1: Re-arrange the equation to have all terms on one side:
  $x^3 - x^2 - 2x = 0$.
• Step 2: Factor out the greatest common factor (GCF), which is $x$:
  $x(x^2 - x - 2) = 0$.
• Step 3: Factor the quadratic expression $x^2 - x - 2$:
  The factors of $-2$ that add up to $-1$ are $-2$ and $1$. Thus, $x^2 - x - 2 = (x-2)(x+1)$.
• Step 4: Combine the factored terms:
  $x(x-2)(x+1) = 0$.

Each factor can be set to zero to find the solutions:

• $x = 0$.
• $x - 2 = 0$, so $x = 2$.
• $x + 1 = 0$, so $x = -1$.

The solutions to the equation are $x = 0, -1, 2$.

Therefore, the correct choice from the given options is:
$x = 0, -1, 2$.

Proceed to solve the given equation:

$y^2+4y+2=-2$

First, let's arrange the equation by moving terms:

$y^2+4y+2=-2 \\ y^2+4y+2+2=0 \\ y^2+4y+4=0$

Note that the expression on the left side can be factored using the perfect square trinomial formula:

$(\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$

As shown below:

$4=2^2$

Therefore, we'll represent the rightmost term as a squared term:

$y^2+4y+4=0 \\ \downarrow\\ \textcolor{red}{y}^2+4y+\textcolor{blue}{2}^2=0$

Now let's examine again the perfect square trinomial formula mentioned earlier:

$(\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

And the expression on the left side in the equation that we obtained in the last step:

$\textcolor{red}{y}^2+\underline{4y}+\textcolor{blue}{2}^2=0$

Note that the terms $\textcolor{red}{y}^2,\hspace{6pt}\textcolor{blue}{2}^2$indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor the expression in question (which is on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined):

$(\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

In other words - we'll ask if we can represent the expression on the left side of the equation as:

$\textcolor{red}{y}^2+\underline{4y}+\textcolor{blue}{2}^2 =0 \\ \updownarrow\text{?}\\ \textcolor{red}{y}^2+\underline{2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{2}}+\textcolor{blue}{2}^2 =0$

And indeed it is true that:

$2\cdot y\cdot2=4y$

Therefore we can represent the expression on the left side of the equation as a perfect square trinomial:

$\textcolor{red}{y}^2+2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{2}+\textcolor{blue}{2}^2=0 \\ \downarrow\\ (\textcolor{red}{y}+\textcolor{blue}{2})^2=0$

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable:

$(y+2)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ y+2=\pm0\\ y+2=0\\ \boxed{y=-2}$

Let's summarize the solution of the equation:

$y^2+4y+2=-2 \\ y^2+4y+4=0 \\ \downarrow\\ \textcolor{red}{y}^2+2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{2}+\textcolor{blue}{2}^2=0 \\ \downarrow\\ (\textcolor{red}{y}+\textcolor{blue}{2})^2=0 \\ \downarrow\\ y+2=0\\ \downarrow\\ \boxed{y=-2}$

Therefore the correct answer is answer D.

To solve the given cubic equation $x^3 - 7x^2 + 6x = 0$, follow these steps:

• Step 1: Identify that the equation can be factored by its Greatest Common Factor (GCF).

There is an $x$ common in all terms: $x(x^2 - 7x + 6) = 0$

• Step 2: Factor the quadratic expression $x^2 - 7x + 6$.

Look for two numbers that multiply to $6$ (the constant term) and add up to $-7$ (the coefficient of the linear term). The numbers are $-1$ and $-6$. Thus:

$x^2 - 7x + 6 = (x - 1)(x - 6)$

• Step 3: Set each factor equal to zero to solve for $x$.

Now that the equation is fully factored as $x(x - 1)(x - 6) = 0$, apply the zero product property:

$x = 0$, $x - 1 = 0$ (so $x = 1$), $x - 6 = 0$ (so $x = 6$)

Thus, the solutions to the equation $x^3 - 7x^2 + 6x = 0$ are $x = 0$, $x = 1$, and $x = 6$.

To solve the equation $x^3 + x^2 - 12x = 0$, follow these steps:

• Step 1: Factor out the greatest common factor. The common factor here is $x$.
• Step 2: The equation becomes $x(x^2 + x - 12) = 0$.
• Step 3: Apply the zero-product property. This gives us two equations to solve: $x = 0$ and $x^2 + x - 12 = 0$.
• Step 4: Solve $x = 0$. This is a straightforward solution: $x = 0$.
• Step 5: Solve the quadratic equation $x^2 + x - 12 = 0$. We will factor it:
• Factor as $(x - 3)(x + 4) = 0$.
• Set each factor equal to zero: $x - 3 = 0$ or $x + 4 = 0$.
• Solving these, we obtain $x = 3$ and $x = -4$.

Therefore, the solutions to the equation are $x = 0$, $x = 3$, and $x = -4$.

Thus, the complete solution set for $x$ is $x = 0, 3, -4$.