Solve {l}{text{Solvetheequation}(x-5)^2-17(x-5)+30=0text{.Whatisthest}}{{llll}{text{(A)}22}&{text{(B)}25}&{text{(C)}27}&{text{(D)}30}}

$$\left. \begin{array} { l } { \text { Solve the equation } ( x - 5 ) ^ { 2 } - 17 ( x - 5 ) + 30 = 0 \text { . What is the st } } \\ { \left. \begin{array} { l l l l } { \text { (A) } 22 } & { \text { (B) } 25 } & { \text { (C) } 27 } & { \text { (D) } 30 } \end{array} \right. } \end{array} \right.$$

Solve for x

$$x=7$$

$$x=20$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-10x+25-17\left(x-5\right)+30=0$$

Use the distributive property to multiply -17 by x-5.

$$x^{2}-10x+25-17x+85+30=0$$

Combine -10x and -17x to get -27x.

$$x^{2}-27x+25+85+30=0$$

Add 25 and 85 to get 110.

$$x^{2}-27x+110+30=0$$

Add 110 and 30 to get 140.

$$x^{2}-27x+140=0$$

To solve the equation, factor x^{2}-27x+140 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-27 ab=140$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 140.

$$-1,-140 -2,-70 -4,-35 -5,-28 -7,-20 -10,-14$$

Calculate the sum for each pair.

$$-1-140=-141 -2-70=-72 -4-35=-39 -5-28=-33 -7-20=-27 -10-14=-24$$

The solution is the pair that gives sum -27.

$$a=-20 b=-7$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-20\right)\left(x-7\right)$$

To find equation solutions, solve x-20=0 and x-7=0.

$$x=20 x=7$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-10x+25-17\left(x-5\right)+30=0$$

Use the distributive property to multiply -17 by x-5.

$$x^{2}-10x+25-17x+85+30=0$$

Combine -10x and -17x to get -27x.

$$x^{2}-27x+25+85+30=0$$

Add 25 and 85 to get 110.

$$x^{2}-27x+110+30=0$$

Add 110 and 30 to get 140.

$$x^{2}-27x+140=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+140. To find a and b, set up a system to be solved.

$$a+b=-27 ab=1\times 140=140$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 140.

$$-1,-140 -2,-70 -4,-35 -5,-28 -7,-20 -10,-14$$

Calculate the sum for each pair.

$$-1-140=-141 -2-70=-72 -4-35=-39 -5-28=-33 -7-20=-27 -10-14=-24$$

The solution is the pair that gives sum -27.

$$a=-20 b=-7$$

Rewrite x^{2}-27x+140 as \left(x^{2}-20x\right)+\left(-7x+140\right).

$$\left(x^{2}-20x\right)+\left(-7x+140\right)$$

Factor out x in the first and -7 in the second group.

$$x\left(x-20\right)-7\left(x-20\right)$$

Factor out common term x-20 by using distributive property.

$$\left(x-20\right)\left(x-7\right)$$

To find equation solutions, solve x-20=0 and x-7=0.

$$x=20 x=7$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-10x+25-17\left(x-5\right)+30=0$$

Use the distributive property to multiply -17 by x-5.

$$x^{2}-10x+25-17x+85+30=0$$

Combine -10x and -17x to get -27x.

$$x^{2}-27x+25+85+30=0$$

Add 25 and 85 to get 110.

$$x^{2}-27x+110+30=0$$

Add 110 and 30 to get 140.

$$x^{2}-27x+140=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -27 for b, and 140 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-27\right)±\sqrt{\left(-27\right)^{2}-4\times 140}}{2}$$

Square -27.

$$x=\frac{-\left(-27\right)±\sqrt{729-4\times 140}}{2}$$

Multiply -4 times 140.

$$x=\frac{-\left(-27\right)±\sqrt{729-560}}{2}$$

Add 729 to -560.

$$x=\frac{-\left(-27\right)±\sqrt{169}}{2}$$

Take the square root of 169.

$$x=\frac{-\left(-27\right)±13}{2}$$

The opposite of -27 is 27.

$$x=\frac{27±13}{2}$$

Now solve the equation x=\frac{27±13}{2} when ± is plus. Add 27 to 13.

$$x=\frac{40}{2}$$

Divide 40 by 2.

$$x=20$$

Now solve the equation x=\frac{27±13}{2} when ± is minus. Subtract 13 from 27.

$$x=\frac{14}{2}$$

Divide 14 by 2.

$$x=7$$

The equation is now solved.

$$x=20 x=7$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-10x+25-17\left(x-5\right)+30=0$$

Use the distributive property to multiply -17 by x-5.

$$x^{2}-10x+25-17x+85+30=0$$

Combine -10x and -17x to get -27x.

$$x^{2}-27x+25+85+30=0$$

Add 25 and 85 to get 110.

$$x^{2}-27x+110+30=0$$

Add 110 and 30 to get 140.

$$x^{2}-27x+140=0$$

Subtract 140 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-27x=-140$$

Divide -27, the coefficient of the x term, by 2 to get -\frac{27}{2}. Then add the square of -\frac{27}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-27x+\left(-\frac{27}{2}\right)^{2}=-140+\left(-\frac{27}{2}\right)^{2}$$

Square -\frac{27}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-27x+\frac{729}{4}=-140+\frac{729}{4}$$

Add -140 to \frac{729}{4}.

$$x^{2}-27x+\frac{729}{4}=\frac{169}{4}$$

Factor x^{2}-27x+\frac{729}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{27}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{27}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$x-\frac{27}{2}=\frac{13}{2} x-\frac{27}{2}=-\frac{13}{2}$$

Add \frac{27}{2} to both sides of the equation.

$$x=20 x=7$$