Solve left(2x+3right)left(5x-7right)^2left(x-1right)^2=0

$$\left( 2x+3 \right) { \left(5x-7 \right) }^{ 2 } { \left(x-1 \right) }^{ 2 } = 0$$

Solve for x

$$x = \frac{7}{5} = 1\frac{2}{5} = 1.4$$

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=1$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-7\right)^{2}.

$$\left(2x+3\right)\left(25x^{2}-70x+49\right)\left(x-1\right)^{2}=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$\left(2x+3\right)\left(25x^{2}-70x+49\right)\left(x^{2}-2x+1\right)=0$$

Use the distributive property to multiply 2x+3 by 25x^{2}-70x+49 and combine like terms.

$$\left(50x^{3}-65x^{2}-112x+147\right)\left(x^{2}-2x+1\right)=0$$

Use the distributive property to multiply 50x^{3}-65x^{2}-112x+147 by x^{2}-2x+1 and combine like terms.

$$50x^{5}-165x^{4}+68x^{3}+306x^{2}-406x+147=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 147 and q divides the leading coefficient 50. List all candidates \frac{p}{q}.

$$±\frac{147}{50},±\frac{147}{25},±\frac{147}{10},±\frac{147}{5},±\frac{147}{2},±147,±\frac{49}{50},±\frac{49}{25},±\frac{49}{10},±\frac{49}{5},±\frac{49}{2},±49,±\frac{21}{50},±\frac{21}{25},±\frac{21}{10},±\frac{21}{5},±\frac{21}{2},±21,±\frac{7}{50},±\frac{7}{25},±\frac{7}{10},±\frac{7}{5},±\frac{7}{2},±7,±\frac{3}{50},±\frac{3}{25},±\frac{3}{10},±\frac{3}{5},±\frac{3}{2},±3,±\frac{1}{50},±\frac{1}{25},±\frac{1}{10},±\frac{1}{5},±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 50x^{5}-165x^{4}+68x^{3}+306x^{2}-406x+147 by x-1 to get 50x^{4}-115x^{3}-47x^{2}+259x-147. Solve the equation where the result equals to 0.

$$50x^{4}-115x^{3}-47x^{2}+259x-147=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -147 and q divides the leading coefficient 50. List all candidates \frac{p}{q}.

$$±\frac{147}{50},±\frac{147}{25},±\frac{147}{10},±\frac{147}{5},±\frac{147}{2},±147,±\frac{49}{50},±\frac{49}{25},±\frac{49}{10},±\frac{49}{5},±\frac{49}{2},±49,±\frac{21}{50},±\frac{21}{25},±\frac{21}{10},±\frac{21}{5},±\frac{21}{2},±21,±\frac{7}{50},±\frac{7}{25},±\frac{7}{10},±\frac{7}{5},±\frac{7}{2},±7,±\frac{3}{50},±\frac{3}{25},±\frac{3}{10},±\frac{3}{5},±\frac{3}{2},±3,±\frac{1}{50},±\frac{1}{25},±\frac{1}{10},±\frac{1}{5},±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 50x^{4}-115x^{3}-47x^{2}+259x-147 by x-1 to get 50x^{3}-65x^{2}-112x+147. Solve the equation where the result equals to 0.

$$50x^{3}-65x^{2}-112x+147=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 147 and q divides the leading coefficient 50. List all candidates \frac{p}{q}.

$$±\frac{147}{50},±\frac{147}{25},±\frac{147}{10},±\frac{147}{5},±\frac{147}{2},±147,±\frac{49}{50},±\frac{49}{25},±\frac{49}{10},±\frac{49}{5},±\frac{49}{2},±49,±\frac{21}{50},±\frac{21}{25},±\frac{21}{10},±\frac{21}{5},±\frac{21}{2},±21,±\frac{7}{50},±\frac{7}{25},±\frac{7}{10},±\frac{7}{5},±\frac{7}{2},±7,±\frac{3}{50},±\frac{3}{25},±\frac{3}{10},±\frac{3}{5},±\frac{3}{2},±3,±\frac{1}{50},±\frac{1}{25},±\frac{1}{10},±\frac{1}{5},±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=\frac{7}{5}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 50x^{3}-65x^{2}-112x+147 by 5\left(x-\frac{7}{5}\right)=5x-7 to get 10x^{2}+x-21. Solve the equation where the result equals to 0.

$$10x^{2}+x-21=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 10 for a, 1 for b, and -21 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 10\left(-21\right)}}{2\times 10}$$

Do the calculations.

$$x=\frac{-1±29}{20}$$

Solve the equation 10x^{2}+x-21=0 when ± is plus and when ± is minus.

$$x=-\frac{3}{2} x=\frac{7}{5}$$

List all found solutions.

$$x=1 x=\frac{7}{5} x=-\frac{3}{2}$$