How to find the square of a sum - PSAT Math

Card 0 of 35

Simplify the radical.

$\sqrt{3283}$

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We can break the square root down into 2 roots of 67 and 49. 49 is a perfect square and reduces to 7.

Expand:

$\left ( 10x+ $\frac{1}{5}\right $)^{2}$$

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Use the perfect square trinomial pattern, setting $A = 10x , B = $\frac{1}{5}$$ :

$(A + $B)^{2}$ = $A^{2}$ + 2AB + B ^{2}$

$\left ( 10x + $\frac{1}{5}$ \right $)^{2}$ = $(10x)^{2}$ + 2 (10x) \left ( $\frac{1}{5}$ \right ) + \left ( $\frac{1}{5}$ \right ) ^{2}$

$= $100x^{2}$ + 4x + $\frac{1}{25}$$

Expand:

$\left ( 10x+ 0.8\right $)^{2}$$

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Use the perfect square trinomial pattern, setting :

$(A + $B)^{2}$ = $A^{2}$ + 2AB + B ^{2}$

$(10x + $0.8)^{2}$ = \left (10x \right $)^{2}$ + 2 \cdot 10x \cdot 0.8 + $0.8^{2}$$

$(10x + $0.8)^{2}$ = $100x^{2}$ + 16x + 0.64$

If $\left (5 $x^{2 }$ - x + 1\right $)^{2}$$ is expanded, what is the coefficient of ?

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$\left (5 $x^{2 }$ - x + 1\right $)^{2}$$

$= \left (5 $x^{2 }$ - x + 1\right ) \left (5 $x^{2 }$ - x + 1\right )$

$= 5 $x^{2 }$ \left (5 $x^{2 }$ - x + 1\right ) - x \left (5 $x^{2 }$ - x + 1\right ) + 1 \left (5 $x^{2 }$ - x + 1\right )$

$= 5 $x^{2 }$ \cdot 5 $x^{2 }$ - 5 $x^{2 }$ \cdot x + $5x^{2 }$ \cdot 1 - x \cdot 5 $x^{2 }$+ x \cdot x - x \cdot 1+ 1\cdot 5 $x^{2 }$ - 1 \cdot x + 1\cdot 1$

$= 25 $x^{4 }$ - 5 $x^{3 }$ + $5x^{2 }$ - 5 $x^{3 }$+ x ^{2 } - x + 5 $x^{2 }$ - x + 1$

$= 25 $x^{4 }$ - 5 $x^{3 }$ - 5 $x^{3 }$ + $5x^{2 }$+ 5 $x^{2 }$ + x ^{2 } - x - x + 1$

$= 25 $x^{4 }$ -10 $x^{3 }$ + 11 $x^{2 }$ -2 x + 1$

The coefficient of is therefore 11.

If $\left (5 $x^{2 }$ - x + 1\right $)^{2}$$ is expanded, what is the coefficient of ?

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$\left (5 $x^{2 }$ - x + 1\right $)^{2}$$

$= \left (5 $x^{2 }$ - x + 1\right ) \left (5 $x^{2 }$ - x + 1\right )$

$= 5 $x^{2 }$ \left (5 $x^{2 }$ - x + 1\right ) - x \left (5 $x^{2 }$ - x + 1\right ) + 1 \left (5 $x^{2 }$ - x + 1\right )$

$= 5 $x^{2 }$ \cdot 5 $x^{2 }$ - 5 $x^{2 }$ \cdot x + $5x^{2 }$ \cdot 1 - x \cdot 5 $x^{2 }$+ x \cdot x - x \cdot 1+ 1\cdot 5 $x^{2 }$ - 1 \cdot x + 1\cdot 1$

$= 25 $x^{4 }$ - 5 $x^{3 }$ + $5x^{2 }$ - 5 $x^{3 }$+ x ^{2 } - x + 5 $x^{2 }$ - x + 1$

$= 25 $x^{4 }$ - 5 $x^{3 }$ - 5 $x^{3 }$ + $5x^{2 }$+ 5 $x^{2 }$ + x ^{2 } - x - x + 1$

$= 25 $x^{4 }$ -10 $x^{3 }$ + 11 $x^{2 }$ -2 x + 1$

The coefficient of is therefore 10.

Simplify the radical.

$\sqrt{3283}$

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We can break the square root down into 2 roots of 67 and 49. 49 is a perfect square and reduces to 7.

Expand:

$\left ( 10x+ $\frac{1}{5}\right $)^{2}$$

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Use the perfect square trinomial pattern, setting $A = 10x , B = $\frac{1}{5}$$ :

$(A + $B)^{2}$ = $A^{2}$ + 2AB + B ^{2}$

$\left ( 10x + $\frac{1}{5}$ \right $)^{2}$ = $(10x)^{2}$ + 2 (10x) \left ( $\frac{1}{5}$ \right ) + \left ( $\frac{1}{5}$ \right ) ^{2}$

$= $100x^{2}$ + 4x + $\frac{1}{25}$$

Expand:

$\left ( 10x+ 0.8\right $)^{2}$$

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Use the perfect square trinomial pattern, setting :

$(A + $B)^{2}$ = $A^{2}$ + 2AB + B ^{2}$

$(10x + $0.8)^{2}$ = \left (10x \right $)^{2}$ + 2 \cdot 10x \cdot 0.8 + $0.8^{2}$$

$(10x + $0.8)^{2}$ = $100x^{2}$ + 16x + 0.64$

If $\left (5 $x^{2 }$ - x + 1\right $)^{2}$$ is expanded, what is the coefficient of ?

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$\left (5 $x^{2 }$ - x + 1\right $)^{2}$$

$= \left (5 $x^{2 }$ - x + 1\right ) \left (5 $x^{2 }$ - x + 1\right )$

$= 5 $x^{2 }$ \left (5 $x^{2 }$ - x + 1\right ) - x \left (5 $x^{2 }$ - x + 1\right ) + 1 \left (5 $x^{2 }$ - x + 1\right )$

$= 5 $x^{2 }$ \cdot 5 $x^{2 }$ - 5 $x^{2 }$ \cdot x + $5x^{2 }$ \cdot 1 - x \cdot 5 $x^{2 }$+ x \cdot x - x \cdot 1+ 1\cdot 5 $x^{2 }$ - 1 \cdot x + 1\cdot 1$

$= 25 $x^{4 }$ - 5 $x^{3 }$ + $5x^{2 }$ - 5 $x^{3 }$+ x ^{2 } - x + 5 $x^{2 }$ - x + 1$

$= 25 $x^{4 }$ - 5 $x^{3 }$ - 5 $x^{3 }$ + $5x^{2 }$+ 5 $x^{2 }$ + x ^{2 } - x - x + 1$

$= 25 $x^{4 }$ -10 $x^{3 }$ + 11 $x^{2 }$ -2 x + 1$

The coefficient of is therefore 11.

If $\left (5 $x^{2 }$ - x + 1\right $)^{2}$$ is expanded, what is the coefficient of ?

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$\left (5 $x^{2 }$ - x + 1\right $)^{2}$$

$= \left (5 $x^{2 }$ - x + 1\right ) \left (5 $x^{2 }$ - x + 1\right )$

$= 5 $x^{2 }$ \left (5 $x^{2 }$ - x + 1\right ) - x \left (5 $x^{2 }$ - x + 1\right ) + 1 \left (5 $x^{2 }$ - x + 1\right )$

$= 5 $x^{2 }$ \cdot 5 $x^{2 }$ - 5 $x^{2 }$ \cdot x + $5x^{2 }$ \cdot 1 - x \cdot 5 $x^{2 }$+ x \cdot x - x \cdot 1+ 1\cdot 5 $x^{2 }$ - 1 \cdot x + 1\cdot 1$

$= 25 $x^{4 }$ - 5 $x^{3 }$ + $5x^{2 }$ - 5 $x^{3 }$+ x ^{2 } - x + 5 $x^{2 }$ - x + 1$

$= 25 $x^{4 }$ - 5 $x^{3 }$ - 5 $x^{3 }$ + $5x^{2 }$+ 5 $x^{2 }$ + x ^{2 } - x - x + 1$

$= 25 $x^{4 }$ -10 $x^{3 }$ + 11 $x^{2 }$ -2 x + 1$

The coefficient of is therefore 10.

Simplify the radical.

$\sqrt{3283}$

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We can break the square root down into 2 roots of 67 and 49. 49 is a perfect square and reduces to 7.

Expand:

$\left ( 10x+ $\frac{1}{5}\right $)^{2}$$

Tap to see back →

Use the perfect square trinomial pattern, setting $A = 10x , B = $\frac{1}{5}$$ :

$(A + $B)^{2}$ = $A^{2}$ + 2AB + B ^{2}$

$\left ( 10x + $\frac{1}{5}$ \right $)^{2}$ = $(10x)^{2}$ + 2 (10x) \left ( $\frac{1}{5}$ \right ) + \left ( $\frac{1}{5}$ \right ) ^{2}$

$= $100x^{2}$ + 4x + $\frac{1}{25}$$

Expand:

$\left ( 10x+ 0.8\right $)^{2}$$

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Use the perfect square trinomial pattern, setting :

$(A + $B)^{2}$ = $A^{2}$ + 2AB + B ^{2}$

$(10x + $0.8)^{2}$ = \left (10x \right $)^{2}$ + 2 \cdot 10x \cdot 0.8 + $0.8^{2}$$

$(10x + $0.8)^{2}$ = $100x^{2}$ + 16x + 0.64$

If $\left (5 $x^{2 }$ - x + 1\right $)^{2}$$ is expanded, what is the coefficient of ?

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$\left (5 $x^{2 }$ - x + 1\right $)^{2}$$

$= \left (5 $x^{2 }$ - x + 1\right ) \left (5 $x^{2 }$ - x + 1\right )$

$= 5 $x^{2 }$ \left (5 $x^{2 }$ - x + 1\right ) - x \left (5 $x^{2 }$ - x + 1\right ) + 1 \left (5 $x^{2 }$ - x + 1\right )$

$= 5 $x^{2 }$ \cdot 5 $x^{2 }$ - 5 $x^{2 }$ \cdot x + $5x^{2 }$ \cdot 1 - x \cdot 5 $x^{2 }$+ x \cdot x - x \cdot 1+ 1\cdot 5 $x^{2 }$ - 1 \cdot x + 1\cdot 1$

$= 25 $x^{4 }$ - 5 $x^{3 }$ + $5x^{2 }$ - 5 $x^{3 }$+ x ^{2 } - x + 5 $x^{2 }$ - x + 1$

$= 25 $x^{4 }$ - 5 $x^{3 }$ - 5 $x^{3 }$ + $5x^{2 }$+ 5 $x^{2 }$ + x ^{2 } - x - x + 1$

$= 25 $x^{4 }$ -10 $x^{3 }$ + 11 $x^{2 }$ -2 x + 1$

The coefficient of is therefore 11.

If $\left (5 $x^{2 }$ - x + 1\right $)^{2}$$ is expanded, what is the coefficient of ?

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$\left (5 $x^{2 }$ - x + 1\right $)^{2}$$

$= \left (5 $x^{2 }$ - x + 1\right ) \left (5 $x^{2 }$ - x + 1\right )$

$= 5 $x^{2 }$ \left (5 $x^{2 }$ - x + 1\right ) - x \left (5 $x^{2 }$ - x + 1\right ) + 1 \left (5 $x^{2 }$ - x + 1\right )$

$= 5 $x^{2 }$ \cdot 5 $x^{2 }$ - 5 $x^{2 }$ \cdot x + $5x^{2 }$ \cdot 1 - x \cdot 5 $x^{2 }$+ x \cdot x - x \cdot 1+ 1\cdot 5 $x^{2 }$ - 1 \cdot x + 1\cdot 1$

$= 25 $x^{4 }$ - 5 $x^{3 }$ + $5x^{2 }$ - 5 $x^{3 }$+ x ^{2 } - x + 5 $x^{2 }$ - x + 1$

$= 25 $x^{4 }$ - 5 $x^{3 }$ - 5 $x^{3 }$ + $5x^{2 }$+ 5 $x^{2 }$ + x ^{2 } - x - x + 1$

$= 25 $x^{4 }$ -10 $x^{3 }$ + 11 $x^{2 }$ -2 x + 1$

The coefficient of is therefore 10.

Simplify the radical.

$\sqrt{3283}$

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We can break the square root down into 2 roots of 67 and 49. 49 is a perfect square and reduces to 7.

Expand:

$\left ( 10x+ $\frac{1}{5}\right $)^{2}$$

Tap to see back →

Use the perfect square trinomial pattern, setting $A = 10x , B = $\frac{1}{5}$$ :

$(A + $B)^{2}$ = $A^{2}$ + 2AB + B ^{2}$

$\left ( 10x + $\frac{1}{5}$ \right $)^{2}$ = $(10x)^{2}$ + 2 (10x) \left ( $\frac{1}{5}$ \right ) + \left ( $\frac{1}{5}$ \right ) ^{2}$

$= $100x^{2}$ + 4x + $\frac{1}{25}$$

Expand:

$\left ( 10x+ 0.8\right $)^{2}$$

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Use the perfect square trinomial pattern, setting :

$(A + $B)^{2}$ = $A^{2}$ + 2AB + B ^{2}$

$(10x + $0.8)^{2}$ = \left (10x \right $)^{2}$ + 2 \cdot 10x \cdot 0.8 + $0.8^{2}$$

$(10x + $0.8)^{2}$ = $100x^{2}$ + 16x + 0.64$

If $\left (5 $x^{2 }$ - x + 1\right $)^{2}$$ is expanded, what is the coefficient of ?

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$\left (5 $x^{2 }$ - x + 1\right $)^{2}$$

$= \left (5 $x^{2 }$ - x + 1\right ) \left (5 $x^{2 }$ - x + 1\right )$

$= 5 $x^{2 }$ \left (5 $x^{2 }$ - x + 1\right ) - x \left (5 $x^{2 }$ - x + 1\right ) + 1 \left (5 $x^{2 }$ - x + 1\right )$

$= 5 $x^{2 }$ \cdot 5 $x^{2 }$ - 5 $x^{2 }$ \cdot x + $5x^{2 }$ \cdot 1 - x \cdot 5 $x^{2 }$+ x \cdot x - x \cdot 1+ 1\cdot 5 $x^{2 }$ - 1 \cdot x + 1\cdot 1$

$= 25 $x^{4 }$ - 5 $x^{3 }$ + $5x^{2 }$ - 5 $x^{3 }$+ x ^{2 } - x + 5 $x^{2 }$ - x + 1$

$= 25 $x^{4 }$ - 5 $x^{3 }$ - 5 $x^{3 }$ + $5x^{2 }$+ 5 $x^{2 }$ + x ^{2 } - x - x + 1$

$= 25 $x^{4 }$ -10 $x^{3 }$ + 11 $x^{2 }$ -2 x + 1$

The coefficient of is therefore 11.

If $\left (5 $x^{2 }$ - x + 1\right $)^{2}$$ is expanded, what is the coefficient of ?

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$\left (5 $x^{2 }$ - x + 1\right $)^{2}$$

$= \left (5 $x^{2 }$ - x + 1\right ) \left (5 $x^{2 }$ - x + 1\right )$

$= 5 $x^{2 }$ \left (5 $x^{2 }$ - x + 1\right ) - x \left (5 $x^{2 }$ - x + 1\right ) + 1 \left (5 $x^{2 }$ - x + 1\right )$

$= 5 $x^{2 }$ \cdot 5 $x^{2 }$ - 5 $x^{2 }$ \cdot x + $5x^{2 }$ \cdot 1 - x \cdot 5 $x^{2 }$+ x \cdot x - x \cdot 1+ 1\cdot 5 $x^{2 }$ - 1 \cdot x + 1\cdot 1$

$= 25 $x^{4 }$ - 5 $x^{3 }$ + $5x^{2 }$ - 5 $x^{3 }$+ x ^{2 } - x + 5 $x^{2 }$ - x + 1$

$= 25 $x^{4 }$ - 5 $x^{3 }$ - 5 $x^{3 }$ + $5x^{2 }$+ 5 $x^{2 }$ + x ^{2 } - x - x + 1$

$= 25 $x^{4 }$ -10 $x^{3 }$ + 11 $x^{2 }$ -2 x + 1$

The coefficient of is therefore 10.